Question

A metal block of heat capcity $80\text{J}{/}^{\circ }\text{C}$ placed in a room at ${20}^{\circ }\text{C}$ is heated electrically. The heater is switched off when the temperature reaches ${30}^{\circ }\text{C}$. The temperature of the block rises at the rate of $2^{\circ }\text{C/s}$ when the heater is switched on and falls at the rate of ${0.2}^{\circ }\text{C/s}$ just after the heater is switched off. Assume Newton's law of cooling to be applicable. If the power radiated at ${25}^{\circ }\text{C}$ represents the average value in the heating process, the time for which the heater was kept on is:

• A. $3.3\text{sec}$
• B. $4.3\text{sec}$
• C. $5.3\text{sec}$
• D. $6.3\text{sec}$

Answer 1

The correct option is C $5.3\text{sec}$

Rate of change of temperature just after the heater is switched on is $\frac{dT}{dt}=2^{\circ }\text{C/s}$
So, the power of the heater or power inident on the metal block, $P_i=\frac{dQ}{dt}=c\frac{dT}{dt}$
$\Rightarrow P_i=80\times 2=160\text{W}$

Similarly, power radiated by the metal block at ${25}^{\circ }\text{C}$,
$P_r=c\frac{dT}{dt}=c\alpha (T-T_0)$
[From Newton's Law of cooling]
$\Rightarrow P_r=80\times (25-20)\times \alpha$
$\Rightarrow P_r=400\alpha ---\left(1\right)$

Also, $\frac{dT}{dt}=\alpha (T-T_0)$
$\Rightarrow 0.2=\alpha (30-20)$
[Just after the heater of switched off]
$\Rightarrow \alpha =0.02--\left(2\right)$

From (1) and (2),
$P_r=400\times 0.02=8\text{W}$

Suppose the time for which the heater was kept on is $t$.
We know that,
Energy incident = Energy radiated + Energy gained
$\Rightarrow E_i=E_r+E_{gain}$
$\Rightarrow 160t=8t+80\times 10[\because P=\frac{E}{t}]$
$\Rightarrow 152t=800$
$\Rightarrow t=5.3\text{s}$