4PH1

Changes of State

Solids, Liquids & Gases · 2 question types

Practise

Exam Frequency Analysis

Past paper frequency (2018 to 2024)

This topic accounts for approximately 6% of your exam marks.

stable
Low
Stable6%

Specific heat capacity, specific latent heat and heating/cooling curves tested consistently.

Definition

  • The (c) of a substance is the amount of energy needed to raise the temperature of 1 kg of the substance by 1 °C
  • The SI unit is joules per kilogram per degree Celsius (J/(kg °C)) (equivalent to J/(kg K) since temperature differences are the same in °C and K)
  • Reference values for a feel:
    • water: 4200 J/(kg °C), very high, which is why oceans moderate climate and water-cooled radiators work
    • aluminium: 900 J/(kg °C)
    • copper: 385 J/(kg °C)
    • iron: 460 J/(kg °C)
    • air: 1000 J/(kg °C)

Implications of high vs low c

  • A substance with a high specific heat capacity:
    • heats up slowly for a given energy input
    • cools down slowly when energy is removed
    • is good for storage heaters, hot-water bottles and household central-heating systems (water carries lots of heat per kilogram)
  • A substance with a low specific heat capacity:
    • heats up quickly and cools down quickly
    • is good for cooking pans and engine pistons (metals heat fast, conduct fast, cool fast)
A copper block, an aluminium block and a beaker of water compared by specific heat capacity: the metals have low values so they warm up and cool down quickly, while water has a high value so it warms and cools slowly
Source: Specific Heat Capacity by Save My Exams

The energy equation

ΔQ = m × c × ΔT

  • where:
    • ΔQ = energy supplied to (or removed from) the substance (J)
    • m = mass of the substance (kg)
    • c = specific heat capacity (J/(kg °C))
    • ΔT = change in temperature (°C)
  • This equation only applies while the substance is not changing state. During a change of state, the temperature does not change so ΔT = 0 and the equation gives zero, but real energy is still being delivered (into the potential store)
Worked example

Calculating thermal energy using ΔQ = m × c × ΔT

A 2.0 kg iron pan is heated from 20 °C to 120 °C. The specific heat capacity of iron is 460 J/(kg °C). Calculate the thermal energy transferred to the pan.

Solution:

  • Identify the temperature change: ΔT = 120 − 20 = 100 °C
  • Write the equation: ΔQ = m × c × ΔT
  • Substitute: ΔQ = 2.0 × 460 × 100
  • = 92 000 J
Worked example

Rearranging ΔQ = m × c × ΔT for an unknown

A 2.0 kg block of metal absorbs 90 000 J of energy and its temperature rises from 20 °C to 70 °C. Calculate the specific heat capacity of the metal.

Solution:

  • Work out the temperature change first: ΔT = 70 − 20 = 50 °C
  • Rearrange ΔQ = m × c × ΔT to make c the subject: c = ΔQ ÷ (m × ΔT)
  • Substitute and evaluate: c = 90 000 ÷ (2.0 × 50) = 90 000 ÷ 100 = 900 J/kg°C

Calculate ΔT as a subtraction before substituting — using the final temperature (70) instead of the change (50) loses the ΔT mark. Show the rearrangement as its own line of working, since the mark scheme credits it separately.