Question

A metal cylinder of mass 0.5 kg is heated electrically by a 12W heater in a room at ${15}^{\circ }C$. The cylinder temperature rises uniformly to ${25}^{\circ }C$ in 5 min and finally becomes constant at ${45}^{\circ }C$. 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 ${20}^{\circ }C$ is 2 W
• B. the rate of loss of heat of the cylinder to surrounding at ${45}^{\circ }C$ is 12 W
• C. Specific heat capacity of metal is $\frac{240}{\left(\frac{3}{2}\right)}J/k{g}^{\circ }C$
• D. none of the above

Answer 1

Answer: (A), (B), (C)

The correct options are
A the rate of loss of heat of the cylinder to surrounding at ${20}^{\circ }C$ is 2 W
B the rate of loss of heat of the cylinder to surrounding at ${45}^{\circ }C$ is 12 W
C Specific heat capacity of metal is $\frac{240}{\left(\frac{3}{2}\right)}J/k{g}^{\circ }C$

$ms\frac{dT}{dt}=12-K(T-15)$
$ATT-{45}^{\circ }C,\frac{dT}{dt}=0$
$\Rightarrow$ Rate of heat loss = K(45 –15) = 12 W
$OrK=\frac{12}{30}=\frac{2}{5}W{/}^{\circ }C$
$AtT={20}^{\circ }C,rateofheatloss=\left(\frac{2}{5}\right)(20-15)=2W$
$ms{\int }_{15}^{25}\frac{dT}{12-\frac{2}{5}(T-15)}=ms{\int }_{15}^{25}\frac{5dT}{90-2T}={\int }_0^{60}dt$
On solving, we get
$\Rightarrow s=\frac{240}{In\left(\frac{3}{2}\right)}J/k{g}^{\circ }C$