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

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The rate of loss of heat of the cylinder to surrounding at 20ºC is 2 W

The rate of loss of heat of the cylinder to surrounding at 45ºC is 2 W

Specific heat capacity of metal is $\frac{240}{\ln (3 / 2)} J / {kg}^{\circ} C$

None of these

answer is A, C.

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Detailed Solution

In final state, $\frac{{dQ}_{e}}{dt} = k (45 - 15) = 12$

$\begin{array}{l} K = \frac{2}{5}; at 20^{\circ} C, \frac{{dQ}_{e}}{dt} = k (20 - 15) = \frac{2}{5} \times 5 = 2 watt. \\ m . s . \frac{dT}{dt} = 12 - \frac{2}{5} (T - 15) \Rightarrow m . s \cdot \frac{dT}{dt} = 18 - \frac{2}{5} T \\ m . s \cdot \int_{15}^{25} \frac{dT}{18 - \frac{2}{5} T} = \int_{0}^{300} dt \Rightarrow s = \frac{240}{\ln (\frac{3}{2})} J / kg -^{\circ} C \end{array}$