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:

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

Where:

• E = Energy change (J)
• m = Mass (kg)
• c = Specific heat capacity (J kg^-1°C^-1)
• $\mathrm{\Delta }\theta$ = 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

$\mathrm{\Delta }\theta$ = 22.9 - 12.1 = 10.8 °C

Step 3: Rearranged specific heat capacity formula

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

Step 4: Substitution and correct evaluation

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

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 ($\mathrm{\Delta }\theta$).
4. Substitute into $E=mc\mathrm{\Delta }\theta$ 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

$E=mc\mathrm{\Delta }\theta$

Step 2: Calculate temperature change

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

Step 3: Substitution and correct evaluation

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

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

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

$20250-135T=840T-21000$

$975T=41250$

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