Question

A metal cylinder of mass 0.5 kg is heated electrically by a 12 W heater in a room at 15°C. The cylinder temperature rises uniformly to 25°C in 5 min and finally becomes constant at 45°C. Assuming that the rate of heat loss is proportional to the excess temperature over the surroundings

• A. The rate of loss of heat of the cylinder to surrounding at 20°C is 2 W
• B. The rate of loss of heat of the cylinder to surrounding at 45°C is 2 W
• C. Specific heat capacity of metal is $\frac{240}{ln(3/2)}$ J/kg°C
• D. None of these

Answer 1

Answer: (A), (C)

A, C

Answer 2

At steady state, the temperature of the cylinder is constant at $T_{steady} = 45^\circ$C. The rate of heat supplied by the heater is equal to the rate of heat lost to the surroundings. Heater power, $P = 12$ W. Ambient temperature, $T_a = 15^\circ$C. The rate of heat loss is proportional to the excess temperature over the surroundings: $\frac{dQ_{loss}}{dt} = k(T - T_a)$, where $k$ is a constant. At steady state: $P = k(T_{steady} - T_a)$ $12 \text{ W} = k(45^\circ\text{C} - 15^\circ\text{C})$ $12 = k(30^\circ\text{C})$ $k = \frac{12}{30} \text{ W/}^\circ\text{C} = 0.4 \text{ W/}^\circ\text{C}$.

Option (A): The rate of loss of heat of the cylinder to the surroundings at $T = 20^\circ$C. Excess temperature $= T - T_a = 20^\circ\text{C} - 15^\circ\text{C} = 5^\circ\text{C}$. Rate of heat loss $= k \times (\text{excess temperature})$ Rate of heat loss $= 0.4 \text{ W/}^\circ\text{C} \times 5^\circ\text{C} = 2 \text{ W}$. Option (A) is correct.

Option (B): The rate of loss of heat of the cylinder to the surroundings at $T = 45^\circ$C. Excess temperature $= T - T_a = 45^\circ\text{C} - 15^\circ\text{C} = 30^\circ\text{C}$. Rate of heat loss $= k \times (\text{excess temperature})$ Rate of heat loss $= 0.4 \text{ W/}^\circ\text{C} \times 30^\circ\text{C} = 12 \text{ W}$. Option (B) states the rate is 2 W, which is incorrect.

Option (C): To find the specific heat capacity ($c$), we use the differential equation for heat transfer: $m \cdot c \cdot \frac{dT}{dt} = P - k(T - T_a)$ Let $\theta = T - T_a$. Then $\frac{d\theta}{dt} = \frac{dT}{dt}$. $m \cdot c \cdot \frac{d\theta}{dt} = P - k\theta$ The cylinder temperature rises from $T_1 = 15^\circ$C to $T_2 = 25^\circ$C in $\Delta t = 5$ min $= 300$ s. Assuming the cylinder was initially at room temperature, $T(0) = 15^\circ$C. So, initial excess temperature $\theta(0) = 15^\circ\text{C} - 15^\circ\text{C} = 0^\circ\text{C}$. At $t=300$ s, the temperature is $T(300) = 25^\circ$C. So, excess temperature at $t=300$ s is $\theta(300) = 25^\circ\text{C} - 15^\circ\text{C} = 10^\circ\text{C}$.

The integrated form of the differential equation $mc \frac{d\theta}{dt} = P - k\theta$ is: $\theta(t) = \frac{P}{k} - \left(\frac{P}{k} - \theta(0)\right) e^{-\frac{k}{mc}t}$ We know $\frac{P}{k} = \frac{12 \text{ W}}{0.4 \text{ W/}^\circ\text{C}} = 30^\circ\text{C}$ (this is the steady-state excess temperature). So, $\theta(t) = 30 - (30 - \theta(0)) e^{-\frac{k}{mc}t}$. With $\theta(0) = 0$: $\theta(t) = 30 - 30 e^{-\frac{k}{mc}t} = 30 (1 - e^{-\frac{k}{mc}t})$. At $t=300$ s, $\theta(300) = 10^\circ$C: $10 = 30 (1 - e^{-\frac{k}{mc} \times 300})$ $\frac{10}{30} = 1 - e^{-\frac{k}{mc} \times 300}$ $\frac{1}{3} = 1 - e^{-\frac{k}{mc} \times 300}$ $e^{-\frac{k}{mc} \times 300} = 1 - \frac{1}{3} = \frac{2}{3}$. Substitute $k=0.4$ W/$^\circ$C and $m=0.5$ kg: $e^{-\frac{0.4}{0.5 \times c} \times 300} = \frac{2}{3}$ $e^{-\frac{0.8}{c} \times 300} = \frac{2}{3}$ $e^{-\frac{240}{c}} = \frac{2}{3}$. Taking the natural logarithm on both sides: $-\frac{240}{c} = \ln\left(\frac{2}{3}\right)$ $-\frac{240}{c} = - \ln\left(\frac{3}{2}\right)$ $\frac{240}{c} = \ln\left(\frac{3}{2}\right)$ $c = \frac{240}{\ln(3/2)}$ J/kg°C. Option (C) is correct.