A metal cylinder of mass 0.5 kg is heated electrically by a 12W heater in a room at $15^{o} C$ . The cylinder temperature rises uniformly to $25^{o} C$ in 5 min and finally becomes constant at $45^{o} C$ . 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 $45^{o} C$ is 12 W

Specific heat capacity of metal is $\frac{240}{\ln (\frac{3}{2})} J / k g^{0} C$

none of the above

the rate of loss of heat of the cylinder to surrounding at $20^{o} C$ is 2 W

answer is A, B, C.

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Detailed Solution

$m s \frac{d T}{d t} = 12 - K (T - 15) A t T = 45^{0} C, d T / d t = 0$

$\Rightarrow$ Rate of heat loss
$\Rightarrow$ Rate of heat loss $= K (45 - 15) = 12 W$
or $K = \frac{12}{30} = \frac{2}{5} W /^{o} C$
At $T = 20^{0} C,$ , rate of heat loss = (2/5) (20 15) = 2W
$m s \int_{15}^{25} \frac{d T}{12 - \frac{2}{5} (T - 15)} = \int_{15}^{25} \frac{5 d T}{90 - 2 T} = \int_{0}^{60} d t$
On solving, we get
$\Rightarrow s = \frac{240}{\ln (3 / 2)} J / k g^{0} C$