(a) Define specific heat capacity. (b)(i) With the aid of a labelled diagram, describe an experiment to determine the specific heat capacity of copper using...

(a) Specific heat capacity is the quantity of heat required to raise the temperature of unit mass (1 kg) of a substance by one kelvin (1 K). Its unit is \(\text{J kg}^{-1}\,\text{K}^{-1}\).

(b)(i) Experiment to determine the specific heat capacity of copper (method of mixtures)

Procedure:

• Weigh the empty copper calorimeter together with its stirrer; record the mass \(m_c\).
• Half-fill the calorimeter with water, weigh again, and obtain the mass of water \(m_w\) by subtraction. Note the initial temperature of the water and calorimeter, \(\theta_1\).
• Weigh the copper ball; record its mass \(m_b\). Suspend it by a thread inside a beaker of water and heat until the water boils steadily. Record the steady temperature of the ball, \(\theta\) (about \(100^\circ\text{C}\)).
• Shake off the adhering water and quickly transfer the hot ball into the calorimeter. Stir gently and record the highest steady temperature of the mixture, \(\theta_2\).

Since heat losses are negligible, heat lost by the copper ball = heat gained by the water + heat gained by the calorimeter and stirrer:

\[ m_b c_b(\theta-\theta_2) = m_w c_w(\theta_2-\theta_1) + m_c c_b(\theta_2-\theta_1) \]

Making \(c_b\), the specific heat capacity of copper, the subject:

\[ c_b = \frac{m_w c_w(\theta_2-\theta_1)}{m_b(\theta-\theta_2) - m_c(\theta_2-\theta_1)} \]

(b)(ii) Two precautions

• Transfer the hot copper ball to the calorimeter quickly and shake off any adhering water first, to minimise heat loss to the surroundings.
• Lag (insulate) the calorimeter and stir the water gently before reading the highest steady temperature.

(c) Final steady temperature of the mixture

Data: copper block \(m_b=24\text{ g}=0.024\text{ kg}\) at \(230^\circ\text{C}\); calorimeter \(m_c=60\text{ g}=0.060\text{ kg}\); water \(m_w=54\text{ g}=0.054\text{ kg}\) at \(31^\circ\text{C}\); \(c_w=4200\ \text{J kg}^{-1}\text{K}^{-1}\); \(c_b=400\ \text{J kg}^{-1}\text{K}^{-1}\). Let the final steady temperature be \(\theta\).

Heat lost by copper block = heat gained by water + heat gained by copper calorimeter:

\[ (0.024)(400)(230-\theta) = (0.054)(4200)(\theta-31) + (0.060)(400)(\theta-31) \] \[ 9.6(230-\theta) = (226.8 + 24)(\theta-31) = 250.8(\theta-31) \] \[ 2208 - 9.6\theta = 250.8\theta - 7774.8 \] \[ 2208 + 7774.8 = 250.8\theta + 9.6\theta \] \[ 9982.8 = 260.4\,\theta \] \[ \theta = \frac{9982.8}{260.4} = 38.34^\circ\text{C} \]

The final steady temperature of the mixture is \(\theta \approx 38.3^\circ\text{C}\).