A2 · Practice questions · Specific heat and latent heat

Warming uses c; changing state uses L.

Six original Cambridge-style questions on specific heat capacity, specific latent heat, heating and cooling curves, and multi-stage energy calculations with Q = mcΔθ and Q = mL.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Each stage separately, then add.

Specific heat capacity: the energy per unit mass needed to raise the temperature of a substance by one kelvin, so Q = mcΔθ, in J kg⁻¹ K⁻¹; use it only while the temperature is changing.
Specific latent heat: the energy per unit mass needed to change the state of a substance at constant temperature, so Q = mL, in J kg⁻¹; fusion for solid to liquid, vaporisation for liquid to gas.
Energy at a plateau: during a change of state the supplied energy increases the potential energy of the molecules (bonds loosened or broken), not their kinetic energy, so the temperature stays constant.

Analysis

[2 marks]

Distinguish between the specific heat capacity and the specific latent heat of a substance, and state the SI unit of each.

Specific heat capacity is the energy required per unit mass to raise the temperature of the substance by one kelvin (one degree), with unit J kg⁻¹ K⁻¹. ✓
Specific latent heat is the energy required per unit mass to change the state of the substance without any change of temperature, with unit J kg⁻¹. ✓

Calculation

[3 marks]

An electric kettle rated at 1.5 kW contains 0.90 kg of water at 15 °C. Calculate the energy needed to bring the water to 100 °C and the minimum time this takes. The specific heat capacity of water is 4200 J kg⁻¹ K⁻¹.

Q = mcΔθ = 0.90 × 4200 × (100 − 15) = 0.90 × 4200 × 85 = 3.2 × 10⁵ J. ✓
t = Q / P = 3.21 × 10⁵ / 1500 = 2.1 × 10² s (214 s, about 3.6 minutes). ✓
This is a minimum because some of the supplied energy warms the kettle itself and is lost to the surroundings, so in practice the time is longer. ✓

Calculation

[3 marks]

A scald from steam is more severe than one from the same mass of boiling water. Calculate the energy released when 25 g of steam at 100 °C condenses on skin and the resulting water then cools to 37 °C, and compare it with the energy released by 25 g of water cooling from 100 °C to 37 °C. Specific latent heat of vaporisation of water = 2.26 × 10⁶ J kg⁻¹, specific heat capacity of water = 4200 J kg⁻¹ K⁻¹.

Condensing: Q = mL = 0.025 × 2.26 × 10⁶ = 5.65 × 10⁴ J. ✓
Cooling the condensed water: Q = mcΔθ = 0.025 × 4200 × (100 − 37) = 6.6 × 10³ J, so the steam releases 5.65 × 10⁴ + 0.66 × 10⁴ = 6.3 × 10⁴ J in total. ✓
Boiling water alone releases only the 6.6 × 10³ J of the cooling stage, about a tenth as much; the latent heat is what makes the steam scald so much worse. ✓

Analysis

[3 marks]

Molten wax in a test tube is allowed to cool in air and its temperature is recorded every minute. The cooling curve falls steadily, then stays flat for several minutes, then falls again. Explain what is happening during the flat section, in terms of the state of the wax and the energy of its molecules.

During the flat section the wax is changing state from liquid to solid (freezing) at its melting point, so its temperature is constant. ✓
The wax is still losing thermal energy to the surroundings, but this energy comes from a decrease in the potential energy of the molecules as they bond into the solid, not from their kinetic energy. ✓
Because the mean kinetic energy of the molecules is unchanged the temperature does not fall; the energy given out per kilogram during this stage is the specific latent heat of fusion. ✓

Calculation

[3 marks]

Ice of mass 0.040 kg at 0 °C is added to 0.30 kg of water at 22 °C in an insulated cup of negligible heat capacity. All the ice melts. Calculate the final temperature of the water. Specific latent heat of fusion of ice = 3.3 × 10⁵ J kg⁻¹, specific heat capacity of water = 4200 J kg⁻¹ K⁻¹.

Energy to melt the ice: Q = mL = 0.040 × 3.3 × 10⁵ = 1.32 × 10⁴ J. ✓
Energy lost by the warm water equals the energy to melt the ice plus the energy to warm the melt water from 0 °C to θ: 0.30 × 4200 × (22 − θ) = 1.32 × 10⁴ + 0.040 × 4200 × θ. ✓
27 720 − 1260θ = 13 200 + 168θ, so 1428θ = 14 520 and θ = 10 °C. ✓

Calculation

[4 marks]

An immersion heater of power 2.0 kW is used to heat 0.50 kg of ethanol, initially at 20 °C, in an open insulated vessel. Ethanol has a specific heat capacity of 2400 J kg⁻¹ K⁻¹ and boils at 78 °C.

(a) Calculate the energy needed to bring the ethanol to its boiling point. (b) Calculate the time taken to reach the boiling point. (c) Once boiling, 0.20 kg of ethanol boils away in 84 s. Calculate the specific latent heat of vaporisation of ethanol given by these data. (d) Explain whether the value in (c) is an overestimate or an underestimate of the true value.

(a) Q = mcΔθ = 0.50 × 2400 × (78 − 20) = 7.0 × 10⁴ J. ✓
(b) t = Q / P = 6.96 × 10⁴ / 2000 = 35 s. ✓
(c) Energy supplied while boiling = Pt = 2000 × 84 = 1.68 × 10⁵ J, so L = Q / m = 1.68 × 10⁵ / 0.20 = 8.4 × 10⁵ J kg⁻¹. ✓
(d) An overestimate: some of the supplied energy is lost to the surroundings rather than vaporising ethanol, so the energy credited to each kilogram vaporised is too large. ✓

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