Specific heat capacity and specific latent heat flashcards

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Specific heat capacity, c
The energy needed to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). Unit J kg⁻¹ K⁻¹.
Specific latent heat, L
The energy needed to change the state of 1 kg of a substance without any change in temperature. Unit J kg⁻¹.
Specific latent heat of fusion
Energy per kilogram to change a substance from solid to liquid (or released liquid to solid) at its melting point.
Specific latent heat of vaporisation
Energy per kilogram to change a substance from liquid to gas (or released gas to liquid) at its boiling point.
Calibration curve (thermistor sensor)
A graph of the sensor's output p.d. against temperature measured with a thermometer; it converts later output readings into temperatures.
Thermostat
A device that switches a heater (or cooler) on and off to keep a temperature near a set value.
Energy for a temperature change
ΔE = mcΔθ (m in kg, c in J kg⁻¹ K⁻¹, Δθ in K or °C). No change of state. (Yes)
Energy for a change of state
ΔE = LΔm (L in J kg⁻¹, Δm = mass that changes state, in kg). (Yes)
Energy from an electrical heater
E = VIt (and P = VI). (Yes)
c from a temperature–time graph
VI = mc × gradient, so c = VI ÷ (m × gradient). (No — derive it)
Funnel method for L (fusion)
L = VIt ÷ (m_heated − m_control). (No — derive it)
Two-power method for L (vaporisation)
L = (P₂ − P₁)t ÷ (m₂ − m₁); the heat-loss rate cancels. (No — derive it)
Heating curve at constant power: what do the sections show?
Sloping sections: one state warming (gradient = P/mc). Flat sections: changes of state at the melting and boiling points.
Heating curve: what does the length of a flat section tell you?
Plateau time × power = mL, so the plateau length is proportional to L. For water the boiling plateau is about 6.8 times the melting one.
Cooling curve of a liquid that solidifies
Falls towards room temperature, levels off at the freezing point while latent heat is released, then falls again once all solid.
θ–t graph for the electrical method: why use the initial gradient?
Early on the block is near room temperature, so losses are negligible; later the graph curves below the tangent as losses grow.
Thermistor calibration curve shape (output across the fixed resistor)
Output p.d. rises with temperature along an S-shaped curve; steepest (most sensitive) where the thermistor's resistance ≈ the fixed resistance.
Circuit for the electrical method for c
Heater in the block, ammeter in series with it, voltmeter in parallel across the heater only, d.c. supply and switch; thermometer in a second hole.
Why is temperature constant during a change of state?
The energy increases the potential energy of the molecules (work against intermolecular forces); mean kinetic energy, and so temperature, is unchanged.
Why is L of vaporisation much larger than L of fusion?
Boiling separates the molecules completely (and the gas pushes back the atmosphere); melting only loosens their arrangement.
Effect of heat losses on a measured c
Δθ is smaller than it should be, so c = E/(mΔθ) comes out too large.
Purpose of the control funnel
It measures ice melted by energy from the surroundings; subtracting its mass leaves the mass melted by the heater alone.
Why use two powers when boiling water to find L?
The heat-loss rate is the same at both (water stays at 100 °C), so subtracting the two energy equations removes it.
Method of mixtures principle
In an insulated system, energy lost by the hot part = energy gained by the cold part; they end at the same final temperature.
Why does the output across the fixed resistor rise as a thermistor warms?
The NTC thermistor's resistance falls, so its share of the supply p.d. falls and the fixed resistor's share rises.
Calibrating a thermistor sensor: key steps
Thermistor beside a thermometer in a stirred water bath; heat in steps; wait for steady readings; record θ and V_out; plot V_out against θ.
Energy to heat 0.50 kg of water from 20 °C to 80 °C? (c = 4180 J kg⁻¹ K⁻¹)
ΔE = 0.50 × 4180 × 60 = 1.25 × 10⁵ J.
A 500 W heater melts 0.20 kg of ice at 0 °C. Minimum time? (L = 3.34 × 10⁵ J kg⁻¹)
t = mL/P = 0.20 × 3.34 × 10⁵ ÷ 500 = 134 s.
24.0 W heater, 0.50 kg block, initial gradient 0.060 K s⁻¹. Specific heat capacity?
c = P ÷ (m × gradient) = 24.0 ÷ (0.50 × 0.060) = 800 J kg⁻¹ K⁻¹.
Funnel method: 12.0 V, 2.0 A for 300 s; heated funnel 30.0 g, control 8.4 g. L?
E = 7200 J; Δm = 21.6 g; L = 7200 ÷ 0.0216 = 3.33 × 10⁵ J kg⁻¹.
Boiling at 50.0 W loses 3.94 g in 200 s; at 90.0 W loses 7.48 g in 200 s. L?
L = 40.0 × 200 ÷ 3.54 × 10⁻³ = 2.26 × 10⁶ J kg⁻¹.
0.100 kg of water at 90 °C is mixed with 0.400 kg of water at 15 °C. Final temperature?
c cancels: θ = (0.100 × 90 + 0.400 × 15) ÷ 0.500 = 30 °C.
Thermostat: 5.0 V supply; switches when 3.0 V is across R; thermistor is 4.0 kΩ at the set temperature. R?
Thermistor takes 2.0 V, so R ÷ 4.0 = 3.0 ÷ 2.0 → R = 6.0 kΩ.
Does Δθ need converting to kelvin?
No: a temperature change has the same value in K and °C. Only absolute temperatures need +273.
Ice that melts and then warms: how many energy terms?
Two: LΔm to melt it, then mcΔθ to warm the meltwater from 0 °C to the final temperature.
Masses given in grams
Convert to kg before using ΔE = mcΔθ or ΔE = LΔm; an answer 1000 times too large or small is the sign you forgot.
Ice straight from a freezer in the funnel method
Some heater energy warms the ice to 0 °C, so less melts per joule and the calculated L is too large. Use ice that is already melting.
After calibrating a thermistor sensor, what must not change?
The supply p.d. and the resistor setting; changing either moves the whole calibration curve.

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