Specific heat capacity

Specific heat capacity (c) is the amount of energy required to raise the temperature of 1 kg of a substance by 1 K or 1°C.

Substances with a higher specific heat capacity need more energy to increase their temperature.

The relationship between energy change, mass, specific heat capacity, and temperature change is given by:

$\text{E = m c }\Delta\theta$

Where:

• E = Energy change (J)
• m = Mass (kg)
• c = Specific heat capacity (J kg^-1°C^-1)
• $\Delta\theta$ = Change in temperature (K or °C)

Worked example - Calculating specific heat capacity

0.250 kg of water is heated from 12.1°C to 22.9°C using a 12 V, 5.30 A heater over 205 s. Calculate the specific heat capacity of water.

Step 1: Calculate heat energy supplied

E = V I t = 12 x 5.30 x 205 = 13,038 J

Step 2: Calculate temperature change

$\Delta\theta$ = 22.9 - 12.1 = 10.8 °C

Step 3: Rearranged specific heat capacity formula

$\text{c = }\frac{E}{m\Delta\theta}$

Step 4: Substitution and correct evaluation

$c = \frac{E}{m\Delta\theta} = \frac{13,038}{0.250 \times 10.8} = 4,828.9 \text{ J kg}^{-1}\text{K}^{-1}$

Determining specific heat capacity experimentally

To find the specific heat capacity of a substance, you need to heat it and measure:

• The energy supplied
• The mass
• The resulting temperature change

Then use these values in the equation above to calculate c.

For a solid:

For a liquid:

Method:

1. Heat the substance, aiming for a temperature rise of about 10 K.
2. Calculate the heating energy (E) using $E = VIt$.
3. Measure the mass (m) and temperature change ($\Delta\theta$).
4. Substitute into $E = mc\Delta\theta$ to calculate c.

Estimating specific heat capacity using the method of mixtures

This method estimates a metal's specific heat capacity:

Steps:

1. Heat a metal block of mass $m_c$ to $T_1$.
2. Transfer it into a calorimeter containing water of mass $m_w$ at $T_0$.
3. Measure the final steady temperature $T$.
4. Heat gained by water = heat lost by metal block $m_c c(T - T_0) = m_w c(T_1 - T)$
5. Rearrange to find c: $c = \frac{m_w c (T_1 - T)}{m_c (T - T_0)}$

Worked example - Calculating specific heat capacity using the method of mixtures

A metal block of mass 0.5 kg is heated to 100°C and then placed in a calorimeter containing 0.4 kg of water at 20°C. The final temperature of the mixture is 25°C. Calculate the specific heat capacity of the metal, assuming the specific heat capacity of water is 4,200 J kg^-1 K^-1.

Step 1: Formula for heat transfer

Heat lost by metal = Heat gained by water

$m_c c_c (T_1 - T) = m_w c_w (T - T_0)$

Where:

• m_c = mass of the metal block (kg)
• c_c = specific heat capacity of the metal (J kg^-1 K^-1)
• T₁ = initial temperature of the metal (°C)
• T = final temperature of the mixture (°C)
• m_w = mass of the water (kg)
• c_w = specific heat capacity of water (J kg^-1 K^-1)
• T₀ = initial temperature of the water (°C)

Step 2: Rearrange to find the specific heat capacity of the metal

$c_c = \frac{m_w c_w (T - T_0)}{m_c (T_1 - T)}$

Step 3: Substitution and correct evaluation

$c_c = \frac{0.4 \times 4200 \times (25 - 20)}{0.5 \times (100 - 25)} = \frac{1680 \times 5}{0.5 \times 75} = \frac{8400}{37.5} = 224 \text{ J kg}^{-1}\text{K}^{-1}$

Worked example - Estimating energy required to heat a substance

Calculate the energy required to heat 0.75 kg of aluminium from 15°C to 75°C. The specific heat capacity of aluminium is 900 J kg^-1 K^-1.

Step 1: Specific heat capacity formula

$E = mc\Delta\theta$

Step 2: Calculate temperature change

$\Delta\theta = 75 - 15 = 60^\circ\text{C}$

Step 3: Substitution and correct evaluation

$E = 0.75 \times 900 \times 60 = 40,500 \text{ J}$

Worked example - Determining final temperature in a calorimetry experiment

0.3 kg of iron at 150°C is placed in 0.2 kg of water at 25°C in a calorimeter. The specific heat capacity of iron is 450 J/kg·K, and that of water is 4,200 J/kg·K. Calculate the final temperature of the mixture.

Step 1: Formula for heat transfer in a calorimetry experiment

Heat lost by iron = Heat gained by water

m_ironc_iron(T_{initial(iron)} - T) = m_waterc_water(T - T_{initial water})

Step 2: Rearrange to find final temperature (T_final)

$0.3 \times 450 \times (150 - T) = 0.2 \times 4200 \times (T - 25)$

Step 3: Substitution and correct evaluation

$135 \times (150 - T) = 840 \times (T - 25)$

$20250 - 135T = 840T - 21000$

$975T = 41250$

$T = \frac{41250}{975} \approx 42.3^\circ\text{C}$