Question

A beaker is fitted with a heating coil and stirrer and contains 40.0 cm³ of liquid A. When the power dissipated in the heating coil is 4.80 W, the temperature of the contents rises from 15.0°C to 35.0°C in 400 s. The experiment is repeated using 20.0 cm³ of liquid A mixed with 20.0 cm³ of liquid B. It is found that, with a heater power of 4.90 W, the temperature again rises from 15.0°C to 35.0°C in 400 s. The specific heat capacity, s of B (in Jkg⁻¹ K⁻¹) is: (Density of A is 1.60 × 10³ kg m⁻³, Specific heat capacity of A is 8.60 × 10² J kg⁻¹ K, Density of B is 2 × 10³ kg m⁻³)

Answer 1

738

Answer 2

The problem describes two experiments where a liquid or mixture is heated, and the temperature rise over a fixed time interval is measured for a given power input. We need to determine the specific heat capacity of liquid B. We can model the energy balance in each experiment by considering the energy supplied by the heater, the energy absorbed by the liquid(s), and the energy absorbed by the beaker, stirrer, and lost to the surroundings. Let $C_{system}$ be the effective heat capacity of the beaker, stirrer, and accounting for heat losses. We assume $C_{system}$ is constant in both experiments since the setup and temperature range are similar.

The energy supplied by the heater is $Q = P \times t$. This energy is used to increase the temperature of the liquid(s) and the rest of the system (beaker, stirrer, surroundings). The heat absorbed by the liquid(s) is $Q_{liquid} = m s \Delta T$, and the heat absorbed by the rest of the system is $Q_{system} = C_{system} \Delta T$. So, the energy balance equation is $P t = Q_{liquid} + Q_{system} = m s \Delta T + C_{system} \Delta T = (ms + C_{system})\Delta T$.

Let's analyze the first experiment with liquid A only. Volume of liquid A, $V_A = 40.0 \text{ cm}^3 = 40.0 \times 10^{-6} \text{ m}^3$. Density of A, $\rho_A = 1.60 \times 10^3 \text{ kg m}^{-3}$. Mass of liquid A, $m_A = \rho_A V_A = (1.60 \times 10^3 \text{ kg m}^{-3}) \times (40.0 \times 10^{-6} \text{ m}^3) = 0.064 \text{ kg}$. Specific heat capacity of A, $s_A = 8.60 \times 10^2 \text{ J kg}^{-1} \text{ K}^{-1}$. Power, $P_1 = 4.80 \text{ W}$. Temperature rise, $\Delta T = 35.0^\circ\text{C} - 15.0^\circ\text{C} = 20.0^\circ\text{C} = 20.0 \text{ K}$. Time, $t = 400 \text{ s}$.

For the first experiment, the energy balance is: $P_1 t = (m_A s_A + C_{system}) \Delta T$ $4.80 \text{ W} \times 400 \text{ s} = (0.064 \text{ kg} \times 8.60 \times 10^2 \text{ J kg}^{-1} \text{ K}^{-1} + C_{system}) \times 20.0 \text{ K}$ $1920 \text{ J} = (55.04 \text{ J K}^{-1} + C_{system}) \times 20.0 \text{ K}$ $1920 = 1100.8 + 20.0 C_{system}$ $20.0 C_{system} = 1920 - 1100.8 = 819.2$ $C_{system} = \frac{819.2}{20.0} = 40.96 \text{ J K}^{-1}$.

Now, let's analyze the second experiment with the mixture of A and B. Volume of liquid A, $V_{A'} = 20.0 \text{ cm}^3 = 20.0 \times 10^{-6} \text{ m}^3$. Mass of liquid A, $m_{A'} = \rho_A V_{A'} = (1.60 \times 10^3 \text{ kg m}^{-3}) \times (20.0 \times 10^{-6} \text{ m}^3) = 0.032 \text{ kg}$. Volume of liquid B, $V_B = 20.0 \text{ cm}^3 = 20.0 \times 10^{-6} \text{ m}^3$. Density of B, $\rho_B = 2 \times 10^3 \text{ kg m}^{-3}$. Mass of liquid B, $m_B = \rho_B V_B = (2 \times 10^3 \text{ kg m}^{-3}) \times (20.0 \times 10^{-6} \text{ m}^3) = 0.040 \text{ kg}$. Let the specific heat capacity of B be $s_B$. Power, $P_2 = 4.90 \text{ W}$. Temperature rise, $\Delta T = 20.0 \text{ K}$. Time, $t = 400 \text{ s}$.

For the second experiment, the energy balance is: $P_2 t = (m_{A'} s_A + m_B s_B + C_{system}) \Delta T$ $4.90 \text{ W} \times 400 \text{ s} = (0.032 \text{ kg} \times 8.60 \times 10^2 \text{ J kg}^{-1} \text{ K}^{-1} + 0.040 \text{ kg} \times s_B + C_{system}) \times 20.0 \text{ K}$ $1960 \text{ J} = (27.52 \text{ J K}^{-1} + 0.040 s_B + C_{system}) \times 20.0 \text{ K}$ Substitute the value of $C_{system} = 40.96 \text{ J K}^{-1}$: $1960 = (27.52 + 0.040 s_B + 40.96) \times 20.0$ $1960 = (68.48 + 0.040 s_B) \times 20.0$ $\frac{1960}{20.0} = 68.48 + 0.040 s_B$ $98.0 = 68.48 + 0.040 s_B$ $0.040 s_B = 98.0 - 68.48 = 29.52$ $s_B = \frac{29.52}{0.040} = \frac{29.52}{40 \times 10^{-3}} = \frac{29520}{40} = \frac{2952}{4} = 738 \text{ J kg}^{-1} \text{ K}^{-1}$.

The specific heat capacity of liquid B is 738 J kg⁻¹ K⁻¹.