Specific heat capacity and specific latent heat
Key idea: H2 Physics lessons on temperature, ideal gases, internal energy and thermodynamic systems.
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The core idea
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Learn the idea
Big question: Why can equal energy inputs cause very different temperature or state changes?
Specific heat capacity relates energy to temperature change, Q = mcΔT. Specific latent heat relates energy to change of state at constant temperature, Q = mL. In experiments, measure electrical energy input and correct for energy transferred to the surroundings or apparatus where possible.
Use heat capacity for a temperature change
Specific heat capacity c is the energy required per unit mass per kelvin of temperature rise: Q = mcΔT. Its unit is J kg⁻¹ K⁻¹. Heat capacity mc belongs to a particular sample; specific heat capacity is a material property under stated conditions.
In electrical experiments, supplied energy is E = VIt or Pt. If energy also warms the container or escapes to the surroundings, treating all of it as energy gained by the sample usually overestimates the sample's c.
Check your understanding: Why measure heating and cooling data near the same temperature?
It helps estimate and correct for the environmental loss rate under similar temperature-difference conditions.
Use latent heat for a phase change
Specific latent heat L is the energy per unit mass needed for a specified phase change at constant temperature: Q = mL. The energy changes microscopic potential energy rather than mean kinetic energy, so temperature ideally remains constant.
A heating curve may contain sloping warming regions and flat phase-change regions. A real plateau can slope because of loss, changing pressure, impurities or uneven temperature; identify which assumption the model makes.
Check your understanding: A 0.060 kg sample melts using 20 kJ. Find L.
L = Q/m = 20 000/0.060 = 3.3 × 10⁵ J kg⁻¹.
Key ideas to keep
- During an ideal phase change, temperature stays constant while internal energy changes.
- Use the mass of the substance undergoing the change.
- A temperature–time gradient depends on power, mass, heat capacity and losses.
See the reasoning
Worked example
Use an energy balance to find an unknown specific heat capacity
Question: A 0.20 kg metal at 200 °C is placed in 0.50 kg water at 20 °C. The final temperature is 30 °C; neglect losses and use cwater = 4200 J kg⁻¹ K⁻¹. Find the metal's specific heat capacity.
Step 1: Find energy gained by the water
Why: The water's mass, c and temperature rise are known.
Working: Qwater = 0.50(4200)(30−20) = 21 000 J.
Step 2: Write the no-loss balance
Why: Neglecting surroundings means energy lost by metal equals energy gained by water.
Working: 0.20c(200−30) = 21 000.
Step 3: Solve and interpret
Why: The metal changes temperature by 170 K.
Working: c = 21 000/[0.20(170)] = 618 J kg⁻¹ K⁻¹.
Answer: Energy gained by water is 0.50(4200)(10) = 21,000 J. The metal loses 0.20c(170), so c = 21,000/[0.20(170)] = 618 J kg⁻¹ K⁻¹.
Check: The metal's c is well below water's, consistent with its much larger temperature change for the same transfer.
Use a hint if needed
Practise with support
Try this
A 0.080 kg ice sample melts at constant temperature using 26.7 kJ. Find its specific latent heat and state which microscopic energy changes.
Hint: Use Q = mL and distinguish kinetic from potential energy.
Check your answer
L = Q/m = 26,700/0.080 = 3.34 × 10⁵ J kg⁻¹. Mean kinetic energy and temperature remain constant; microscopic potential energy increases as the structure changes.
Now work without the hint
Practise independently
Your turn
A heater supplies 36 kJ to 0.60 kg of liquid, raising it by 20 K. A further 90 kJ boils 0.040 kg. Find c and L.
Check your answer
c = 36,000/[0.60(20)] = 3.0 × 10³ J kg⁻¹ K⁻¹. L = 90,000/0.040 = 2.25 × 10⁶ J kg⁻¹.
Avoid these traps
Common mistakes
Common mistake
Specific heat capacity describes the whole object.
What is wrong with this reasoning?
Show better thinking
Heat capacity is for a whole object; specific heat capacity is per unit mass per kelvin.
Common mistake
Latent heating increases temperature throughout a phase change.
What is wrong with this reasoning?
Show better thinking
During a constant-temperature phase change, energy changes microscopic potential energy rather than mean kinetic energy.
Write for the examiner
Exam guidance
For practical questions, identify the largest loss and say how the method or analysis reduces its effect.
Exam-style practice [6 marks]
Define c and L with units, and explain the microscopic difference between warming and melting.
Plan before you answer
- Define c and L with units.
- Match each equation to temperature change or phase change.
- Explain the microscopic energy change.
Mark your answer and compare the model
Marking points
Tick each point only if your answer states it clearly.
Model answer
c is energy per unit mass per kelvin, J kg⁻¹ K⁻¹. L is phase-change energy per unit mass, J kg⁻¹. Warming raises mean kinetic energy; melting at constant temperature primarily raises microscopic potential energy.
Come back in three days
Check what stayed with you
Recall question
Why does a temperature plateau during boiling not imply that no energy is transferred?
Check the answer
Energy is still supplied as mL. It changes microscopic potential energy and phase rather than mean kinetic energy, so thermodynamic temperature remains constant during the phase change.
Syllabus and review details
This lesson covers the listed H2 Physics 9478 outcomes. Temperature and ideal-gas ideas lead into the first law of thermodynamics. Use ΔU = Q + W, where W is work done on the system; for expansion against constant external pressure, work done by the gas is pΔV and work done on the gas is −pΔV.
- GCE A-Level H2 PhysicsTopic 13(g) · 2027Checked against the syllabus · partial topic coverageOfficial 9478 syllabus
Course and syllabus information
- Course
- GCE A-Level H2 Physics
- Edition
- GCE A-Level H2 Physics 2027