Thermal Properties of Matter • Topic 2 of 3

Calorimetry & Change of State

Heat is a form of energy that is transferred between a system and its surroundings because of a temperature difference. Its SI unit is the joule (J), though the calorie is also used: $1\ \text{cal}=4.186\ \text{J}$. When heat is supplied to a body it may raise the body's temperature, or it may change its state without any change in temperature.

Specific heat capacity ($c$) is the amount of heat needed to raise the temperature of unit mass of a substance by one degree. The heat exchanged when a body of mass $m$ changes temperature by $\Delta T$ is:

  • $Q=mc\,\Delta T$, with $c$ measured in $\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$.
  • Water has an unusually high specific heat, $c\approx4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$, which is why it is used as a coolant and why coastal climates are mild.

Molar specific heat ($C$) is the heat needed to raise the temperature of one mole of a substance by one degree: $Q=nC\,\Delta T$, where $n$ is the number of moles. Its unit is $\text{J}\,\text{mol}^{-1}\,\text{K}^{-1}$. For gases there are two values — at constant pressure ($C_p$) and at constant volume ($C_v$) — because a gas can also do work as it expands.

Principle of calorimetry. When two bodies at different temperatures are brought into contact in an insulated container (a calorimeter), heat flows from the hotter to the colder until they reach a common temperature. Assuming no heat is lost to the surroundings, the law of conservation of energy gives:

  • Heat lost by the hot body = Heat gained by the cold body.
  • This is the basis for measuring specific heats and latent heats experimentally.

Latent heat ($L$). During a change of state, the temperature stays constant even though heat is being supplied or removed. The energy goes into breaking or forming the bonds between molecules rather than increasing their kinetic energy. The heat required to change the state of unit mass without any change in temperature is the latent heat:

  • $Q=mL$, with $L$ in $\text{J}\,\text{kg}^{-1}$.
  • Latent heat of fusion ($L_f$): for melting/freezing; for ice $L_f\approx3.34\times10^{5}\ \text{J}\,\text{kg}^{-1}$.
  • Latent heat of vaporisation ($L_v$): for boiling/condensing; for water $L_v\approx22.6\times10^{5}\ \text{J}\,\text{kg}^{-1}$.

The phase-change graph of temperature against heat supplied makes this clear. As ice below $0\,^\circ$C is heated, its temperature rises; at $0\,^\circ$C it stays flat (a plateau) while the ice melts; the temperature then climbs to $100\,^\circ$C, where another, longer plateau marks boiling; only after all the water has turned to steam does the temperature rise again. The flat plateaus correspond to latent heat being absorbed at constant temperature.

Phase-change graph: temperature versus heat supplied for waterHeat supplied (Q)Temp0100melting (L_f)boiling (L_v)icewatersteam
Solved Examples
1
Worked Example
How much heat is required to raise the temperature of 2 kg of water from $20\,^\circ$C to $80\,^\circ$C? ($c_{water}=4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$)
Solution
  1. Step 1: Use $Q=mc\,\Delta T$ with $\Delta T=80-20=60\ \text{K}$.
  2. Step 2: Substitute: $Q=2\times4186\times60$.
  3. Step 3: Compute: $Q=502{,}320\ \text{J}\approx5.02\times10^{5}\ \text{J}$.

Answer: $Q\approx5.02\times10^{5}\ \text{J}$.

2
Worked Example
Find the heat required to convert 0.5 kg of ice at $0\,^\circ$C into water at $0\,^\circ$C. ($L_f=3.34\times10^{5}\ \text{J}\,\text{kg}^{-1}$)
Solution
  1. Step 1: A change of state at constant temperature needs $Q=mL_f$.
  2. Step 2: Substitute: $Q=0.5\times3.34\times10^{5}$.
  3. Step 3: Compute: $Q=1.67\times10^{5}\ \text{J}$.

Answer: $Q=1.67\times10^{5}\ \text{J}$.

3
Worked Example
A 200 g aluminium block at $100\,^\circ$C is dropped into 500 g of water at $20\,^\circ$C. Find the final temperature. ($c_{Al}=900$, $c_{water}=4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$; neglect heat lost to the calorimeter.)
Solution
  1. Step 1: Heat lost by aluminium = heat gained by water: $m_{Al}c_{Al}(100-T)=m_w c_w(T-20)$.
  2. Step 2: Substitute: $0.2\times900\times(100-T)=0.5\times4186\times(T-20)$, i.e. $180(100-T)=2093(T-20)$.
  3. Step 3: Expand: $18000-180T=2093T-41860$, so $59860=2273T$, giving $T\approx26.3\,^\circ\text{C}$.

Answer: Final temperature $\approx26.3\,^\circ\text{C}$.

4
Worked Example
Calculate the heat needed to convert 1 kg of water at $100\,^\circ$C completely into steam at $100\,^\circ$C. ($L_v=22.6\times10^{5}\ \text{J}\,\text{kg}^{-1}$)
Solution
  1. Step 1: At the boiling point the temperature is constant, so $Q=mL_v$.
  2. Step 2: Substitute: $Q=1\times22.6\times10^{5}$.
  3. Step 3: Compute: $Q=2.26\times10^{6}\ \text{J}$.

Answer: $Q=2.26\times10^{6}\ \text{J}$.

5
Worked Example
How much heat is needed to convert 100 g of ice at $0\,^\circ$C into water at $50\,^\circ$C? ($L_f=3.34\times10^{5}$, $c_{water}=4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$)
Solution
  1. Step 1: Two steps: melt the ice, then warm the water. $Q=mL_f+mc\,\Delta T$ with $m=0.1\ \text{kg}$.
  2. Step 2: Melting: $Q_1=0.1\times3.34\times10^{5}=3.34\times10^{4}\ \text{J}$. Warming: $Q_2=0.1\times4186\times50=2.09\times10^{4}\ \text{J}$.
  3. Step 3: Total $Q=Q_1+Q_2=3.34\times10^{4}+2.09\times10^{4}=5.43\times10^{4}\ \text{J}$.

Answer: $Q\approx5.43\times10^{4}\ \text{J}$.

6
Worked Example
A piece of metal of mass 0.4 kg is heated to $200\,^\circ$C and then dropped into 0.2 kg of water at $25\,^\circ$C. The final temperature is $45\,^\circ$C. Find the specific heat of the metal. ($c_{water}=4186\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$)
Solution
  1. Step 1: Heat lost by metal = heat gained by water: $m_m c_m(200-45)=m_w c_w(45-25)$.
  2. Step 2: Substitute: $0.4\times c_m\times155=0.2\times4186\times20$, i.e. $62\,c_m=16744$.
  3. Step 3: Solve: $c_m=\frac{16744}{62}\approx270\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$.

Answer: $c_m\approx270\ \text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$.

Key Points

  • Heat is energy in transit due to a temperature difference; SI unit joule, with $1\ \text{cal}=4.186\ \text{J}$.
  • Specific heat: $Q=mc\,\Delta T$ (unit $\text{J}\,\text{kg}^{-1}\,\text{K}^{-1}$); molar specific heat: $Q=nC\,\Delta T$.
  • Principle of calorimetry: heat lost by the hot body = heat gained by the cold body (no losses).
  • Latent heat: $Q=mL$; temperature stays constant during a change of state.
  • Phase-change graph shows plateaus at melting ($L_f$) and boiling ($L_v$) points where heat is absorbed at constant temperature.
Quick Check — 4 MCQs
Tap an option to check your answer0 / 4
Q1.The heat required to change the temperature of a body is given by:
Explanation: Heating without a change of state follows $Q=mc\,\Delta T$.
Q2.During the melting of ice at $0\,^\circ$C, the temperature of the mixture:
Explanation: Heat supplied during melting goes into latent heat, so the temperature stays at $0\,^\circ$C.
Q3.The SI unit of latent heat is:
Explanation: Latent heat $L=Q/m$ is energy per unit mass, so its unit is $\text{J}\,\text{kg}^{-1}$.
Q4.The principle of calorimetry is a statement of the conservation of:
Explanation: Heat lost equals heat gained, which is conservation of energy in an insulated system.
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