Question

A block of mass 100 g and temperature 0 °C is immersed in a calorimeter containing 500 g water at temperature 45 °C. Specific heat s of the material of the block depends on temperature θ according to equation $s = s_0(1 + k\theta)$, where $s_0$ = 4.2 × $10^3$ J/(kg.° C), k = 0.10 °C-1 and $\theta$ is temperature in degree Celsius. The final temperature (in °C) of water in the calorimeter is 5n, the volume of n is ___. Neglect heat capacity of the calorimeter and loss of heat to the surroundings. Specific heat of water is 4.2 × $10^3$ J/(kg°C).

Answer 1

6

Answer 2

The heat gained by the block is calculated by integrating the specific heat over the temperature change: $Q_{gain} = \int_{0}^{T_f} m_b s_0(1 + k\theta) d\theta = m_b s_0 \left( T_f + \frac{k T_f^2}{2} \right)$

The heat lost by the water is: $Q_{loss} = m_w c_w (45 - T_f)$

Equating $Q_{gain}$ and $Q_{loss}$: $0.1 \times 4.2 \times 10^3 \left( T_f + \frac{0.1 T_f^2}{2} \right) = 0.5 \times 4.2 \times 10^3 (45 - T_f)$ $0.42 (T_f + 0.05 T_f^2) = 2.1 (45 - T_f)$ $T_f + 0.05 T_f^2 = 5 (45 - T_f)$ $0.05 T_f^2 + 6 T_f - 225 = 0$ $T_f^2 + 120 T_f - 4500 = 0$

Solving the quadratic equation gives $T_f = 30$ °C. Given $T_f = 5n$, we have $5n = 30$, so $n = 6$.