ExaminerPrep › Question bank › IAL Physics › 5.1 Specific heat capacity and specific latent heat
Edexcel International A Level Physics · Unit 5 · Thermodynamics, Radiation, Oscillations and Cosmology
40 free, high-quality past-paper-style questions on Specific heat capacity and specific latent heat for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
Which is the SI unit of specific latent heat?
| Answer | Mark | Notes |
|---|---|---|
| From ΔE = LΔm, L = ΔE/Δm, so its unit is J kg−1. There is no kelvin in it because the temperature does not change during a change of state. | ✓ 1 | Answer C |
Answer: C
Exam tip: Derive a unit from the defining equation rather than memorising it: energy ÷ mass for L; energy ÷ (mass × temperature change) for c.
A pan contains 0.60 kg of cooking oil of specific heat capacity 1900 J kg−1 K−1. The oil is heated from 15 °C to 40 °C.
How much energy is transferred to the oil?
| Answer | Mark | Notes |
|---|---|---|
| Δθ = 40 − 15 = 25 K. ΔE = mcΔθ = 0.60 × 1900 × 25 = 2.85 × 104 J. | ✓ 1 | Answer B |
Answer: B
Common error: Adding 273 to a temperature change. A change of 25 °C is a change of 25 K.
A beaker of crushed ice is heated steadily. While the ice is melting, the temperature stays at 0 °C.
Which statement explains this?
| Answer | Mark | Notes |
|---|---|---|
| During a change of state the energy does work against the forces between the molecules, increasing their potential energy. The mean kinetic energy, and so the temperature, does not change. | ✓ 1 | Answer D |
Answer: D
Exam tip: Temperature tracks the mean kinetic energy of the molecules. If the temperature is constant while energy is supplied, the energy must be going into potential energy.
A travel kettle has a heating element of power 2.00 kW. It contains 0.750 kg of water at 20.0 °C. Specific heat capacity of water = 4180 J kg−1 K−1.
State what is meant by the specific heat capacity of a substance.
Calculate the energy needed to bring the water to 100 °C.
Calculate the minimum time this takes.
Give one reason why the actual time is longer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy needed to raise the temperature of 1 kg of the substance by 1 K | ✓ 1 | |
| Part (b) | ||
| Δθ = 80.0 K; ΔE = 0.750 × 4180 × 80.0 | ✓ 1 | |
| = 2.51 × 105 J (250 800 J) | ✓ 1 | |
| Part (c) | ||
| t = 250 800 ÷ 2000 = 125.4 s, so 126 s (rounded up: 125 s is too short) (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| energy is also used to heat the kettle itself / energy is transferred to the surroundings | ✓ 1 | |
Exam tip: Convert kW to W before dividing: 2.00 kW = 2000 W.
A boiling tube holds 8.0 g of molten paraffin wax and is left to cool in the air. The temperature of the wax falls, stays at 54 °C from 4.0 minutes to 19.0 minutes, then falls again. While the temperature is constant, the wax loses energy to the surroundings at an estimated rate of 1.8 W.
State the melting point of the wax.
Explain why the temperature stays constant even though the wax is losing energy.
Estimate the specific latent heat of fusion of the wax.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 54 °C | ✓ 1 | |
| Part (b) | ||
| the wax is solidifying / changing state | ✓ 1 | |
| the energy lost comes from a decrease in the potential energy of the molecules, released as latent heat, so their mean kinetic energy and the temperature do not change | ✓ 1 | |
| Part (c) | ||
| energy released = 1.8 × (15.0 × 60) = 1620 J | ✓ 1 | |
| L = 1620 ÷ 0.0080 = 2.0 × 105 J kg−1 | ✓ 1 | |
Exam tip: Convert the time of the flat section into seconds before multiplying by a power in watts.
A liquid of mass 0.400 kg is heated in an insulated container. The graph shows the temperature of the liquid against the total energy supplied to it.
What is the specific heat capacity of the liquid?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, 40 kJ raises the temperature from 20 °C to 60 °C, so Δθ = 40 K. c = ΔE/(mΔθ) = 40 000 ÷ (0.400 × 40) = 2.5 × 103 J kg−1 K−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: On a temperature–energy graph the gradient is 1/(mc). Read a large triangle from the line, convert kJ to J, then divide by the mass as well.
A solid is heated at a constant rate until it has become a gas. The graph shows how its temperature changes with time.
In which section is the substance entirely liquid while the mean kinetic energy of its molecules increases?
| Answer | Mark | Notes |
|---|---|---|
| QR is melting and ST is boiling: the temperature is constant there, so the mean kinetic energy is constant. PQ is the solid warming. Between the end of melting (R) and the start of boiling (S) the substance is all liquid and its temperature rises, so the answer is RS. | ✓ 1 | Answer C |
Answer: C
Exam tip: On a heating curve, sloping sections are one state warming up (kinetic energy rising); flat sections are changes of state (potential energy rising).
Blocks of four materials, P, Q, R and S, each have a mass of 0.50 kg. Each block is given 9.0 kJ of energy with no losses. The bar chart shows the specific heat capacity of each material.
Which block has a temperature rise of 20 K?
| Answer | Mark | Notes |
|---|---|---|
| c = ΔE/(mΔθ) = 9000 ÷ (0.50 × 20) = 900 J kg−1 K−1. The chart shows 900 J kg−1 K−1 for P. | ✓ 1 | Answer A |
Answer: A
Common error: Forgetting to divide by the mass. Specific heat capacity is per kilogram, so ΔE/Δθ alone is too large or too small unless m = 1 kg.
A 0.150 kg block of ice at −20 °C is placed in an insulated container and heated by a 52.5 W heater. The graph shows the temperature of the contents of the container against time. Assume all the energy from the heater goes to the ice or water.
Use the graph to state the melting point of ice, and explain how the graph shows it.
Use the graph to determine the energy supplied while the ice is melting.
Calculate the specific latent heat of fusion of ice given by these results.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 0 °C: the graph is horizontal at this temperature while the ice melts | ✓ 1 | |
| Part (b) | ||
| flat section from 2.0 min to 18.0 min, so 16.0 min = 960 s | ✓ 1 | |
| E = 52.5 × 960 = 5.04 × 104 J | ✓ 1 | |
| Part (c) | ||
| L = ΔE/Δm = 50 400 ÷ 0.150 | ✓ 1 | |
| = 3.36 × 105 J kg−1 (allow error carried forward) | ✓ 1 | |
Common error: Using the time at the end of the flat section (18 min) instead of its length. The melting time is the difference between the start and end of the flat section.
An aluminium pan of mass 0.500 kg is heated on an electric hotplate. The figure shows the reading of a thermometer attached to the pan at the start and at the end of the heating. Specific heat capacity of aluminium = 900 J kg−1 K−1.
Use the figure to find the rise in temperature of the pan.
Calculate the energy gained by the pan.
The hotplate supplied energy at 250 W for 2.0 minutes. Calculate the efficiency of heating the pan.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 63 − 21 = 42 K | ✓ 1 | |
| Part (b) | ||
| ΔE = 0.500 × 900 × 42 | ✓ 1 | |
| = 1.89 × 104 J (18 900 J) | ✓ 1 | |
| Part (c) | ||
| energy supplied = 250 × 120 = 30 000 J | ✓ 1 | |
| efficiency = 18 900 ÷ 30 000 = 0.63 = 63% (allow error carried forward) | ✓ 1 | |
Common error: Adding 273 to a temperature change. A rise of 42 °C is a rise of 42 K.
A sample is heated at a constant rate with no energy losses. On its temperature–time graph the gradient while it is solid is 0.75 K s−1, and the gradient after it has completely melted is 0.45 K s−1.
What is the ratio (specific heat capacity of the liquid) ÷ (specific heat capacity of the solid)?
| Answer | Mark | Notes |
|---|---|---|
| Power P = mc × gradient with the same P and m, so c ∝ 1/gradient. Ratio = 0.75 ÷ 0.45 = 1.7 (1.67). The liquid warms more slowly, so its specific heat capacity is larger. | ✓ 1 | Answer C |
Answer: C
Exam tip: A smaller gradient on a heating curve at constant power means a larger specific heat capacity.
0.200 kg of water at 80 °C is poured into an insulated flask containing 0.300 kg of water at 20 °C. No energy is transferred to the flask or the surroundings.
What is the final temperature of the water?
| Answer | Mark | Notes |
|---|---|---|
| Energy lost by hot water = energy gained by cold water: 0.200 × c × (80 − θ) = 0.300 × c × (θ − 20). So 16 − 0.2θ = 0.3θ − 6, giving θ = 44 °C. | ✓ 1 | Answer B |
Answer: B
Exam tip: For two samples of the same liquid, c cancels: the final temperature is the mass-weighted mean of the starting temperatures.
In an experiment to find the specific latent heat of fusion of ice, a heater in a funnel of melting ice is supplied with 12.0 V and 2.50 A for 200 s. The heated funnel collects 25.0 g of water. An identical control funnel with no heater collects 7.0 g of water in the same time.
What value does this give for the specific latent heat of fusion?
| Answer | Mark | Notes |
|---|---|---|
| E = VIt = 12.0 × 2.50 × 200 = 6000 J. Ice melted by the heater = 25.0 − 7.0 = 18.0 g = 0.0180 kg. L = 6000 ÷ 0.0180 = 3.33 × 105 J kg−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: Subtract the control mass: the room melts ice in both funnels, and only the extra mass in the heated funnel is due to the heater.
A student finds the specific heat capacity of a metal block using an electrical heater, an ammeter, a voltmeter and a thermometer. The block is not insulated.
How does the energy transferred to the surroundings affect the value the student calculates, and why?
| Answer | Mark | Notes |
|---|---|---|
| The electrical energy VIt is measured correctly, but some of it leaves the block, so Δθ is smaller than it should be. c = VIt/(mΔθ) with Δθ too small gives a value that is too large. | ✓ 1 | Answer A |
Answer: A
Common error: Thinking losses must make every result “too small”. Work out which measured quantity is affected and where it sits in the equation.
A thermostat uses an NTC thermistor in series with a variable resistor across a fixed supply. A switching circuit turns a heater off when the p.d. across the variable resistor rises to a set value. The thermostat is to be adjusted so that the heater switches off at a lower temperature than before.
How should the resistance of the variable resistor be changed, and why?
| Answer | Mark | Notes |
|---|---|---|
| At a lower temperature the NTC thermistor's resistance is higher. For the variable resistor still to take the same share of the supply p.d., its resistance must be increased in proportion. | ✓ 1 | Answer A |
Answer: A
Exam tip: At the switching point the two p.d.s are fixed, so the two resistances must stay in the same ratio. If one goes up, the other must go up too.
A student measures the specific heat capacity of vegetable oil. An immersion heater is placed in 0.400 kg of oil in an insulated beaker. The p.d. across the heater is 12.0 V and the current in it is 2.50 A. The oil is stirred and its temperature recorded every 60 s:
time / s: 0, 60, 120, 180, 240 — temperature / °C: 19.0, 21.3, 23.6, 25.9, 28.2.
State how the ammeter and the voltmeter are connected in the heater circuit.
Use the results to determine the specific heat capacity of the oil.
The beaker loses some energy to the surroundings. Explain how this affects the value obtained.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ammeter in series with the heater | ✓ 1 | |
| voltmeter in parallel with (across) the heater only | ✓ 1 | |
| Part (b) | ||
| rate of temperature rise = (28.2 − 19.0) ÷ 240 = 0.0383 K s−1 | ✓ 1 | |
| power = 12.0 × 2.50 = 30.0 W, and VI = mc × (Δθ/Δt) | ✓ 1 | |
| c = 30.0 ÷ (0.400 × 0.0383) = 1960 J kg−1 K−1 | ✓ 1 | |
| Part (c) | ||
| the temperature rise is smaller than it would be with no losses | ✓ 1 | |
| so c = VIt/(mΔθ) is too large / an overestimate | ✓ 1 | |
Model sketch
Exam tip: When the temperature rises steadily, the rate of rise from the first and last readings equals the gradient of the graph; use the gradient if the points curve.
A test tube of solid candle wax at room temperature is placed in a hot-water bath, which supplies energy to the wax at a steady rate. The temperature of the wax is recorded until it is completely liquid and well above its melting point. The temperature–time graph rises, becomes horizontal for several minutes, then rises again less steeply than at the start.
State what is meant by the specific latent heat of fusion.
Explain the shape of the graph. Your answer should refer to the energy of the molecules.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the energy needed to change 1 kg of a substance from solid to liquid with no change in temperature | ✓ 1 | |
| Part (b) | ||
| Indicative content, one mark for each point: | — | |
| in the first sloping section the energy supplied increases the (mean) kinetic energy of the molecules, so the temperature rises | ✓ 1 | |
| the horizontal section is where the wax is melting / changing state at its melting point | ✓ 1 | |
| during melting the energy supplied increases the potential energy of the molecules / does work against the forces between them | ✓ 1 | |
| the mean kinetic energy is unchanged, so the temperature stays constant | ✓ 1 | |
| once all the wax has melted, the energy again increases the kinetic energy, so the temperature rises | ✓ 1 | |
| the final section is less steep because liquid wax has a larger specific heat capacity than solid wax (rate = P/mc) | ✓ 1 | |
Exam tip: Link each section of the graph to what happens to the molecules' kinetic and potential energy, and say why the gradient changes, not just that it does.
A student calibrates a temperature sensor. An NTC thermistor is connected in series with a 3.3 kΩ resistor across a 5.0 V supply, and the output p.d. is measured across the resistor. Some of her results are:
temperature / °C: 20, 30, 40, 50 — output p.d. / V: 1.75, 2.23, 2.70, 3.13.
Describe how she should obtain the calibration readings.
Explain why the output p.d. increases as the temperature increases.
The sensor later gives an output of 2.40 V. Assuming the calibration is linear between neighbouring readings, estimate the temperature.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| place the thermistor next to a thermometer in a water bath (with the connections kept dry) | ✓ 1 | |
| heat the water in steps, stirring and waiting for steady readings before recording temperature and output p.d. | ✓ 1 | |
| cover a wide range of temperatures (e.g. 0 °C to near 100 °C) and plot output p.d. against temperature | ✓ 1 | |
| Part (b) | ||
| the resistance of the thermistor decreases as its temperature increases | ✓ 1 | |
| so the thermistor takes a smaller share of the 5.0 V and the p.d. across the fixed resistor increases | ✓ 1 | |
| Part (c) | ||
| 2.40 V lies between 2.23 V (30 °C) and 2.70 V (40 °C): 30 + 10 × (2.40 − 2.23) ÷ (2.70 − 2.23) | ✓ 1 | |
| = 34 °C (33.6 °C) | ✓ 1 | |
Model sketch
Exam tip: A calibration curve converts a reading into the quantity you want. Read it, or interpolate between neighbouring points, rather than recalculating from the circuit.
Two identical funnels are filled with crushed melting ice, and each drains into a beaker. A heater is placed in the ice in one funnel only. The heater is switched on for 240 s with a p.d. of 12.0 V and a current of 3.20 A. In this time 36.4 g of water collects below the heated funnel and 8.2 g collects below the other funnel.
Explain the purpose of the funnel without a heater.
Calculate the specific latent heat of fusion of ice given by these results.
In a repeat, the ice is used straight from a freezer at −18 °C. Explain the effect on the value obtained for the specific latent heat.
Suggest one further improvement to the procedure.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it measures the mass of ice melted by energy from the surroundings, which can then be subtracted from the heated funnel's mass | ✓ 1 | |
| Part (b) | ||
| E = 12.0 × 3.20 × 240 = 9216 J | ✓ 1 | |
| mass melted by the heater = 36.4 − 8.2 = 28.2 g | ✓ 1 | |
| L = 9216 ÷ 0.0282 = 3.27 × 105 J kg−1 | ✓ 1 | |
| Part (c) | ||
| some of the heater's energy is used to warm the ice to 0 °C before it can melt | ✓ 1 | |
| so less ice melts for the same energy and the calculated L is too large | ✓ 1 | |
| Part (d) | ||
| make sure the heater is completely covered by ice / wait until water drips at a steady rate from both funnels before starting / use a joulemeter | ✓ 1 | |
Common error: Forgetting to convert grams to kilograms before dividing. An answer of a few hundred J kg−1 is the sign.
The figure shows the circuit used to heat a liquid, with the readings of the meters and the stop-clock at the moment the heater is switched off. The stop-clock was started when the heater was switched on, and the meter readings stayed constant.
How much energy was transferred by the heater?
| Answer | Mark | Notes |
|---|---|---|
| The stop-clock reads 5 minutes 0 seconds, so t = 300 s. E = VIt = 11.8 × 3.40 × 300 = 12 036 J = 1.20 × 104 J. | ✓ 1 | Answer C |
Answer: C
Common error: Using the stop-clock reading in minutes. Convert to seconds before using E = VIt.
A well-insulated metal block of mass 0.800 kg is heated by a 36.0 W heater. The graph shows the temperature of the block against time.
What is the specific heat capacity of the metal?
| Answer | Mark | Notes |
|---|---|---|
| From the graph the temperature rises from 18 °C to 38 °C in 400 s, a gradient of 0.0500 K s−1. P = mc(Δθ/Δt), so c = 36.0 ÷ (0.800 × 0.0500) = 900 J kg−1 K−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: For steady heating, use the gradient of the temperature–time graph: c = P ÷ (m × gradient).
An NTC thermistor is connected in series with a fixed 2.0 kΩ resistor across a 6.0 V supply of negligible internal resistance. A switching circuit connected across the fixed resistor turns on a cooling fan when the p.d. across the fixed resistor reaches 4.0 V. The graph shows how the resistance of the thermistor varies with temperature.
At what temperature does the fan switch on?
| Answer | Mark | Notes |
|---|---|---|
| The thermistor has 6.0 − 4.0 = 2.0 V across it. The current is the same in both, so RT ÷ 2.0 kΩ = 2.0 V ÷ 4.0 V, giving RT = 1.0 kΩ. The graph shows 1.0 kΩ at 60 °C. | ✓ 1 | Answer D |
Answer: D
Exam tip: In a potential divider the p.d.s are in the same ratio as the resistances. Find the resistance first, then read the graph.
A heated glass block of mass 0.250 kg is placed in 0.200 kg of water in an insulated container. The graph shows the temperatures of the glass and of the water against time. Specific heat capacity of water = 4180 J kg−1 K−1. No energy is transferred to the container or the surroundings.
What is the specific heat capacity of the glass?
| Answer | Mark | Notes |
|---|---|---|
| From the graph the glass falls from 90 °C to 30 °C (60 K) and the water rises from 16 °C to 30 °C (14 K). Energy lost = energy gained: 0.250 × c × 60 = 0.200 × 4180 × 14, so c = 11 704 ÷ 15.0 = 780 J kg−1 K−1. | ✓ 1 | Answer A |
Answer: A
Exam tip: In a mixture the energy lost by the hot object equals the energy gained by the cold one. Each object has its own mass and its own temperature change.
A laboratory water bath contains 3.00 kg of water. Its heater is controlled by a timer, and the graph shows the heater power against time. Assume that all the energy goes to the water. Specific heat capacity of water = 4180 J kg−1 K−1.
What is the rise in temperature of the water?
| Answer | Mark | Notes |
|---|---|---|
| Energy = area under the graph = 2000 × 120 + 500 × 240 = 240 000 + 120 000 = 360 000 J. Δθ = 360 000 ÷ (3.00 × 4180) = 28.7 K. | ✓ 1 | Answer B |
Answer: B
Exam tip: When the power changes, the energy is the area under the power–time graph. Add the areas of the separate sections.
A student heats a 1.00 kg copper block with an electric heater. The p.d. across the heater is 12.0 V and the current in it is 3.20 A. The block is not insulated, and at t = 0 it is at room temperature. The graph shows the temperature of the block against time, with a tangent drawn to the curve at t = 0.
Calculate the power of the heater.
Use the tangent to determine the initial rate of rise of temperature of the block.
Calculate the specific heat capacity of copper given by these results.
Explain why the gradient of the graph decreases with time, and why the gradient at t = 0 is used.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| P = VI = 12.0 × 3.20 = 38.4 W | ✓ 1 | |
| Part (b) | ||
| gradient of tangent from two points far apart, e.g. (50 − 20) ÷ (300 − 0) | ✓ 1 | |
| = 0.100 K s−1 (allow 0.095–0.105) | ✓ 1 | |
| Part (c) | ||
| P = mc(Δθ/Δt), so c = 38.4 ÷ (1.00 × 0.100) | ✓ 1 | |
| = 384 J kg−1 K−1 (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| as the block gets hotter than the room, energy is transferred to the surroundings at a greater rate, so less of the heater's power raises the temperature | ✓ 1 | |
| at t = 0 the block is at room temperature, so the loss is negligible and all 38.4 W heats the block | ✓ 1 | |
Exam tip: When a heating curve bends, draw the tangent at the start, where the losses are smallest, and read its gradient from a large triangle.
A student makes a temperature sensor from the circuit in Figure 1(a). The supply has negligible internal resistance. She calibrates the sensor by measuring the output p.d. at a range of known temperatures, and draws the calibration graph in Figure 1(b).
Explain why the output p.d. decreases as the temperature increases.
Later the output p.d. is 2.0 V. Use Figure 1(b) to find the temperature of the thermistor.
Calculate the resistance of the thermistor at this temperature.
Use Figure 1(b) to state the range of temperature over which this sensor is most sensitive. Give a reason for your answer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the resistance of the (NTC) thermistor decreases as its temperature rises | ✓ 1 | |
| so the thermistor takes a smaller share of the 5.0 V supply, and the output across it falls | ✓ 1 | |
| Part (b) | ||
| 50 °C (allow 48–52 °C) | ✓ 1 | |
| Part (c) | ||
| p.d. across the 1.5 kΩ resistor = 5.0 − 2.0 = 3.0 V, so I = 3.0 ÷ 1500 = 2.0 × 10−3 A | ✓ 1 | |
| R = 2.0 ÷ 2.0 × 10−3 = 1.0 kΩ (1000 Ω) | ✓ 1 | |
| Part (d) | ||
| about 20 °C to 40 °C (allow any range within 10–50 °C) | ✓ 1 | |
| the graph is steepest there, so the output p.d. changes most for each kelvin of temperature change | ✓ 1 | |
Exam tip: Sensitivity is the change in output per unit change of input. On a calibration graph it is the gradient, so the steepest part is the most sensitive range.
The figure shows an electric shower. Cold water flows through a chamber containing a heating element and leaves through the shower head. The temperatures and the power of the element are shown on the figure. Specific heat capacity of water = 4180 J kg−1 K−1. Assume all the energy from the element goes to the water.
Calculate the mass of water that flows through the shower each second.
1 kg of water has a volume of 1.0 litre. Calculate the volume of water delivered per minute.
In winter the cold water enters at 6.0 °C and the flow is reduced to 0.0600 kg s−1. Calculate the temperature of the water leaving the shower.
Suggest why the water leaving the shower head is slightly cooler than your calculation predicts.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Δθ = 40 − 12 = 28 K | ✓ 1 | |
| P = (m/t)cΔθ, so m/t = 8500 ÷ (4180 × 28) | ✓ 1 | |
| = 0.073 kg s−1 (0.0726) | ✓ 1 | |
| Part (b) | ||
| 0.0726 × 60 = 4.4 litres (4.36) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| Δθ = 8500 ÷ (0.0600 × 4180) = 33.9 K | ✓ 1 | |
| θ = 6.0 + 33.9 = 39.9 °C (about 40 °C, almost the same as in summer) | ✓ 1 | |
| Part (d) | ||
| some energy is transferred to the chamber, the pipe and the surroundings instead of to the water | ✓ 1 | |
Exam tip: For a flowing liquid, use P = (m/t)cΔθ: the mass flow rate in kg s−1 replaces the mass.
A student measures the specific latent heat of vaporisation of ethanol. Ethanol in an insulated flask is heated by an immersion heater connected to a joulemeter. In each run the ethanol starts at room temperature, and the heater is switched off when a measured mass of ethanol has evaporated. The graph shows the energy supplied against the mass evaporated, with a line of best fit.
Use the graph to determine the specific latent heat of vaporisation of ethanol.
Explain why the gradient gives a better value than dividing the energy by the mass for a single run.
Suggest what the intercept of the graph represents.
State one safety precaution for this experiment.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient from two points far apart on the line, e.g. (0, 1.5 kJ) and (10.0 g, 10.0 kJ) | ✓ 1 | |
| gradient = 8.5 kJ ÷ 10.0 g = 0.85 kJ g−1 | ✓ 1 | |
| = 8.5 × 105 J kg−1 (allow 8.2–8.8 × 105) | ✓ 1 | |
| Part (b) | ||
| every run includes the same extra energy that does not evaporate any ethanol | ✓ 1 | |
| this appears as the intercept and is removed by using the gradient; a single run would give a value that is too large | ✓ 1 | |
| Part (c) | ||
| the energy needed to warm the ethanol (and heater and flask) from room temperature to its boiling point before any evaporates | ✓ 1 | |
| Part (d) | ||
| ethanol is flammable: no naked flames nearby / keep the room ventilated / do not let the heater run dry | ✓ 1 | |
Exam tip: When every run has the same fixed energy cost, plot energy against mass. The gradient is L and the fixed cost drops out as the intercept.
0.050 kg of ice at 0 °C is added to 0.400 kg of water at 30.0 °C in an insulated container. All the ice melts. Specific heat capacity of water = 4180 J kg−1 K−1. Specific latent heat of fusion of ice = 3.34 × 105 J kg−1.
What is the final temperature of the water?
| Answer | Mark | Notes |
|---|---|---|
| Energy lost by the warm water = energy to melt the ice + energy to warm the meltwater: 0.400 × 4180 × (30.0 − θ) = 0.050 × 3.34 × 105 + 0.050 × 4180 × θ. So 50 160 − 1672θ = 16 700 + 209θ, giving θ = 18 °C (17.8 °C). | ✓ 1 | Answer A |
Answer: A
Exam tip: Ice that melts and then warms needs two energy terms: LΔm for melting and mcΔθ for warming the meltwater from 0 °C.
Water is kept boiling by an electric heater in an open container. In 200 s, 2.80 g of water boils away when the heater power is 40.0 W, and 5.45 g boils away when the power is 70.0 W. The rate of energy loss to the surroundings is the same in both runs.
What is the specific latent heat of vaporisation of water given by these results?
| Answer | Mark | Notes |
|---|---|---|
| Subtracting the two runs removes the heat loss: L = (70.0 − 40.0) × 200 ÷ ((5.45 − 2.80) × 10−3) = 6000 ÷ 2.65 × 10−3 = 2.26 × 106 J kg−1. | ✓ 1 | Answer B |
Answer: B
Exam tip: With the same loss rate in both runs, use the difference in power and the difference in mass. A single run always gives a value that is too large.
A student finds the specific heat capacity of brass by the method of mixtures. A 0.120 kg brass weight is heated in boiling water at 100.0 °C and then quickly transferred to 0.150 kg of water at 19.5 °C in a polystyrene cup. The highest temperature reached by the water is 24.8 °C. The thermometer reads to 0.1 °C. Specific heat capacity of water = 4180 J kg−1 K−1.
Calculate the specific heat capacity of brass.
The accepted value is about 380 J kg−1 K−1. Explain one reason why the student's value is smaller.
The uncertainty in each temperature reading is ± 0.05 °C. Calculate the percentage uncertainty in the temperature rise of the water and explain why this temperature measurement, rather than the temperature fall of the brass, most limits the precision of the result.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy gained by water = 0.150 × 4180 × (24.8 − 19.5) = 3323 J | ✓ 1 | |
| energy lost by brass = 0.120 × c × (100.0 − 24.8) | ✓ 1 | |
| c = 3323 ÷ (0.120 × 75.2) = 368 J kg−1 K−1 | ✓ 1 | |
| Part (b) | ||
| either the brass cools during the transfer, so its real temperature drop is less than 75.2 K, or energy is lost from the cup, so the water's measured temperature rise is too small | ✓ 1 | |
| either way the energy given out per kelvin by the brass is underestimated, so the calculated c is too small | ✓ 1 | |
| Part (c) | ||
| uncertainty in a difference = ± 0.1 °C, so 0.1 ÷ 5.3 × 100 = 1.9% | ✓ 1 | |
| the brass's temperature change (75.2 K) has a far smaller percentage uncertainty (about 0.1%), so the small rise of the water dominates | ✓ 1 | |
Exam tip: For a temperature change, add the absolute uncertainties of the two readings before dividing by the change.
A coffee machine heats milk by passing steam at 100 °C into it. The steam condenses in the milk. A jug contains 0.180 kg of milk at 5.0 °C, which is to be heated to 65.0 °C. Specific heat capacity of milk = 3900 J kg−1 K−1. Specific heat capacity of water = 4180 J kg−1 K−1. Specific latent heat of vaporisation of water = 2.26 × 106 J kg−1.
Calculate the energy gained by the milk.
Calculate the mass of steam that must condense in the milk.
Explain why steam heats the milk far more effectively than the same mass of water at 100 °C would.
State one assumption you made in your calculation.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| ΔE = 0.180 × 3900 × (65.0 − 5.0) | ✓ 1 | |
| = 4.21 × 104 J (42 120 J) | ✓ 1 | |
| Part (b) | ||
| each kg of steam releases L on condensing and then 4180 × (100 − 65.0) as the condensed water cools | ✓ 1 | |
| energy per kg = 2.26 × 106 + 146 300 = 2.41 × 106 J | ✓ 1 | |
| m = 42 120 ÷ 2.406 × 106 = 0.0175 kg (17.5 g) (allow error carried forward) | ✓ 1 | |
| Part (c) | ||
| the steam releases its latent heat of vaporisation as it condenses | ✓ 1 | |
| this is about 2.26 × 106 J per kg, much more than the water releases in cooling from 100 °C to 65.0 °C (1.46 × 105 J per kg) | ✓ 1 | |
| Part (d) | ||
| no energy is transferred to the jug or the surroundings / the condensed water mixes with the milk and ends at 65.0 °C | ✓ 1 | |
Exam tip: Condensing steam gives two energy terms: the latent heat released at 100 °C, and the energy released as the condensed water cools to the final temperature.
A greenhouse heater is controlled by a thermostat. An NTC thermistor is connected in series with a variable resistor across a 6.0 V supply of negligible internal resistance. A switching circuit, which draws negligible current, turns the heater off when the p.d. across the variable resistor rises to 4.0 V. The thermistor's resistance is 3.2 kΩ at 30 °C and 5.0 kΩ at 20 °C.
Calculate the resistance of the variable resistor needed for the heater to switch off at 30 °C.
Calculate the p.d. across the variable resistor at 20 °C and state whether the heater is on or off.
Suggest why the switching circuit must draw negligible current.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| p.d. across the thermistor = 6.0 − 4.0 = 2.0 V | ✓ 1 | |
| same current in both, so R ÷ 3.2 = 4.0 ÷ 2.0 | ✓ 1 | |
| R = 6.4 kΩ | ✓ 1 | |
| Part (b) | ||
| V = 6.0 × 6.4 ÷ (6.4 + 5.0) | ✓ 1 | |
| = 3.4 V (3.37 V) (allow error carried forward) | ✓ 1 | |
| below 4.0 V, so the heater is on | ✓ 1 | |
| Part (c) | ||
| otherwise it would act as a resistor in parallel with the variable resistor and lower the output p.d., so the heater would switch at the wrong temperature | ✓ 1 | |
Exam tip: At a switching point, write down both p.d.s first. The resistances are then in the same ratio as the p.d.s.
A container of water stands on a digital balance that reads to 0.01 g. A heater keeps the water boiling. The mass lost in 300 s is measured at two heater powers: 5.84 g at 50.0 W and 9.82 g at 80.0 W. The rate of energy loss to the surroundings is the same at both powers.
Explain why the water must already be boiling steadily before the mass lost is measured.
Calculate the specific latent heat of vaporisation of water.
Calculate the rate of energy loss to the surroundings.
Each mass lost is found from two balance readings, each with an uncertainty of ± 0.01 g. Calculate the percentage uncertainty in the difference 9.82 g − 5.84 g.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| otherwise some of the energy would raise the temperature of the water instead of turning it into steam | ✓ 1 | |
| Part (b) | ||
| (80.0 − 50.0) × 300 = L × (9.82 − 5.84) × 10−3; the loss term cancels | ✓ 1 | |
| 9000 = L × 3.98 × 10−3 | ✓ 1 | |
| L = 2.26 × 106 J kg−1 | ✓ 1 | |
| Part (c) | ||
| 50.0 × 300 = 2.26 × 106 × 5.84 × 10−3 + h × 300 | ✓ 1 | |
| h = 6.0 W (allow error carried forward) | ✓ 1 | |
| Part (d) | ||
| each mass lost has an uncertainty of ± 0.02 g, so the difference has ± 0.04 g | ✓ 1 | |
| 0.04 ÷ 3.98 × 100 = 1.0% | ✓ 1 | |
Exam tip: In a sum or a difference, add the absolute uncertainties. Only then convert to a percentage, using the value of the difference itself.
A vacuum flask of liquid nitrogen stands on a top-pan balance. The nitrogen boils slowly because energy leaks in from the surroundings. At 300 s a 12.0 W heater in the nitrogen is switched on. The graph shows the balance reading against time. The rate at which energy leaks in stays constant.
What value do these results give for the specific latent heat of vaporisation of nitrogen?
| Answer | Mark | Notes |
|---|---|---|
| Before the heater: 6.0 g lost in 300 s = 0.020 g s−1. With the heater: 24 g in 300 s = 0.080 g s−1. The heater causes the extra 0.060 g s−1 = 6.0 × 10−5 kg s−1, so L = 12.0 ÷ 6.0 × 10−5 = 2.0 × 105 J kg−1. | ✓ 1 | Answer C |
Answer: C
Exam tip: When something is already boiling off before you start heating, only the extra rate of mass loss is due to the heater.
A tube of a molten compound cools in the air. The graph is its cooling curve; just before it begins to solidify, the liquid cools at a steady rate. The specific heat capacity of the liquid compound is 2300 J kg−1 K−1. Assume that the rate of energy loss to the surroundings is the same just before and during solidification.
What is the specific latent heat of fusion of the compound?
| Answer | Mark | Notes |
|---|---|---|
| Liquid cooling rate = (82 − 70) ÷ 300 s = 0.040 K s−1, so the rate of energy loss is m × 2300 × 0.040. Solidification lasts 40 − 5 = 35 min = 2100 s, so mL = m × 2300 × 0.040 × 2100 and L = 1.9 × 105 J kg−1 (1.93 × 105). The mass cancels. | ✓ 1 | Answer B |
Answer: B
Exam tip: If the loss rate is the same in two sections, write it as mc × gradient for the sloping section and use it for the flat section. Keep every time in seconds.
In a series of trials, different masses of ice at 0 °C are added to 0.300 kg of water at 40.0 °C in an insulated flask. After stirring, the final temperature of the contents is recorded. The graph shows the final temperature against the mass of ice added. Specific heat capacity of water = 4180 J kg−1 K−1. Assume no energy is transferred to the flask or the surroundings.
Explain why the final temperature is lower when more ice is added.
Use the graph to find the smallest mass of ice that gives a final temperature of 0 °C.
Use your answer to (b) to calculate the specific latent heat of fusion of ice.
Explain why the graph is horizontal for masses of ice greater than this.
Calculate the mass of ice left unmelted when 200 g of ice is added.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the ice needs energy (latent heat) to melt, and the meltwater then needs energy to warm up from 0 °C | ✓ 1 | |
| this energy comes from the warm water, so more ice takes more energy from it and the water cools further | ✓ 1 | |
| Part (b) | ||
| 150 g (allow 145–155 g) | ✓ 1 | |
| Part (c) | ||
| energy given out by the water cooling from 40.0 °C to 0 °C = 0.300 × 4180 × 40.0 = 50 160 J | ✓ 1 | |
| L = 50 160 ÷ 0.150 = 3.34 × 105 J kg−1 (allow 3.24–3.46 × 105 from the range in (b)) | ✓ 1 | |
| Part (d) | ||
| the water has no more energy to give before it reaches 0 °C, so the extra ice does not melt and the ice–water mixture stays at 0 °C | ✓ 1 | |
| Part (e) | ||
| mass melted = 50 160 ÷ 3.34 × 105 = 0.150 kg | ✓ 1 | |
| unmelted = 0.200 − 0.150 = 0.050 kg (50 g) (allow error carried forward) | ✓ 1 | |
Exam tip: At the point where the line just reaches 0 °C, all the ice has melted and the meltwater has not warmed at all, so the only energy terms are mcΔθ for the water and LΔm for the ice.
A student uses one funnel of crushed ice at 0 °C with an electric heater buried in the ice. She does not use a control funnel. For each heater power she waits until water drips steadily, then measures the rate at which meltwater collects below the funnel. The graph shows her results with a line of best fit.
Explain why the line of best fit does not pass through the origin.
Use the graph to determine the specific latent heat of fusion of ice.
Calculate the rate at which energy reaches the ice from the surroundings.
Identify the anomalous result and suggest a reason for it.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| energy from the surroundings also melts ice, so water collects even when the heater power is zero | ✓ 1 | |
| Part (b) | ||
| gradient from two points far apart on the line, e.g. (0, 0.020) and (60, 0.200): 0.180 g s−1 ÷ 60 W | ✓ 1 | |
| = 3.0 × 10−3 g J−1 = 3.0 × 10−6 kg J−1; rate = P/L + constant, so L = 1/gradient | ✓ 1 | |
| L = 1 ÷ 3.0 × 10−6 = 3.3 × 105 J kg−1 (allow 3.2–3.5 × 105) | ✓ 1 | |
| Part (c) | ||
| intercept = 0.020 g s−1 = 2.0 × 10−5 kg s−1 | ✓ 1 | |
| rate = 2.0 × 10−5 × 3.33 × 105 = 6.7 W (allow 6.0–7.3 W; allow error carried forward from (b)) | ✓ 1 | |
| Part (d) | ||
| the result at 40 W, which lies below the line | ✓ 1 | |
| less ice melted than expected for that power, e.g. the ice had melted away from the heater so some energy heated the air and funnel / the reading was started before the water was dripping steadily | ✓ 1 | |
Exam tip: Write the relationship in the form y = mx + c before using a graph: here rate = (1/L) × power + (melting by the surroundings), so the gradient is 1/L, not L.
A student designs a thermostat for a reptile tank. Figure 1(a) shows the circuit: an NTC thermistor and a fixed resistor R in series across a 9.0 V supply of negligible internal resistance. A switching circuit, which draws negligible current, turns a heater off when the output p.d. Vout across R rises to a set value. Figure 1(b) shows how the resistance of the thermistor varies with temperature. The tank should be kept close to 30 °C.
Use Figure 1(b) to find the resistance of the thermistor at 25 °C and at 35 °C.
The student has a 1.0 kΩ resistor and a 3.3 kΩ resistor to use as R. For each resistor, calculate Vout at 25 °C and at 35 °C. Hence state which resistor gives the larger change in Vout between these temperatures.
Explain why the resistor that gives the larger change in Vout makes a better thermostat.
The 3.3 kΩ resistor is used. Use Figure 1(b) to determine the p.d. at which the switching circuit must operate so that the heater switches off at 30 °C.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| 4.0 kΩ at 25 °C and 2.6 kΩ at 35 °C (both needed) | ✓ 1 | |
| Part (b) | ||
| 1.0 kΩ: 9.0 × 1.0 ÷ 5.0 = 1.80 V and 9.0 × 1.0 ÷ 3.6 = 2.50 V, a change of 0.70 V | ✓ 1 | |
| 3.3 kΩ: 9.0 × 3.3 ÷ 7.3 = 4.07 V and 9.0 × 3.3 ÷ 5.9 = 5.03 V, a change of 0.97 V | ✓ 1 | |
| the 3.3 kΩ resistor gives the larger change (allow error carried forward from (a)) | ✓ 1 | |
| Part (c) | ||
| Vout changes more per kelvin, so the switching p.d. is reached over a smaller change in temperature and the heater switches off closer to the set temperature (the sensor is more sensitive) | ✓ 1 | |
| Part (d) | ||
| from Figure 1(b), thermistor resistance at 30 °C = 3.2 kΩ | ✓ 1 | |
| Vout = 9.0 × 3.3 ÷ (3.3 + 3.2) = 4.6 V (4.57 V; allow 4.5–4.7 V) | ✓ 1 | |
Exam tip: A thermostat is most sensitive near its set temperature when the fixed resistor has about the same resistance as the thermistor at that temperature.
Figure 1 shows the apparatus a student uses to determine the specific heat capacity of aluminium.
Describe how the student should use the apparatus in Figure 1 to obtain an accurate value for the specific heat capacity of aluminium. Your answer should include any change you would make to the apparatus as shown, and how the results are processed.
In Figure 1 the thermometer stands loosely in its hole. Explain how this affects the temperature readings and suggest an improvement.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Indicative content, one mark for each point (the thermometer contact is credited in (b), not here): | — | |
| the block in Figure 1 is not insulated: wrap it in insulating material (lagging) to reduce energy transfer to the surroundings | ✓ 1 | |
| record the starting temperature, then switch on the heater and start the stop-clock at the same time | ✓ 1 | |
| record the ammeter and voltmeter readings, checking they stay constant | ✓ 1 | |
| record the temperature at regular intervals, e.g. every 60 s, for a rise of about 10–20 K | ✓ 1 | |
| plot temperature against time and find the gradient of the straight section | ✓ 1 | |
| c = VI ÷ (m × gradient), using the mass m = 1.00 kg given in Figure 1 | ✓ 1 | |
| Part (b) | ||
| air is a poor conductor, so the thermometer lags behind the block and reads lower than the block's temperature | ✓ 1 | |
| put a few drops of oil (or water) in the hole to improve the thermal contact | ✓ 1 | |
Exam tip: In a method description, say what you measure, with which instrument, how you reduce losses, and exactly how the result is calculated from a graph.
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