Question

When a body is placed in surroundings at a constant temperature of $20°C$, and heated by a $10-W$ heater, its temperature remains constant at $40°C$. If the temperature of the body is now raised from $20°C$ to $80°C$ in $5$ minutes at a uniform rate, the total heat it will lose to the surroundings will be

• A. $3000J$
• B. $3600J$
• C. $4500J$
• D. $5400J$

Answer 1

The correct option is C $4500J$

Let $\theta$ be the temperature of the body. It is given that the temperature of the surroundings is ${\theta }_0={20}^{\circ }C$.
Rate of loss of heat, $\frac{dQ}{dt}=k(\theta -{\theta }_0)$
For $\theta ={40}^{\circ }C$,
$\frac{dQ}{dt}=10W=k({40}^{\circ }C-{20}^{\circ }C)$
or $k=\frac{10W}{{20}^{\circ }C}=\frac{1}{2}\frac{W}{{}^{\circ }C}$
When the temperature of the body is raised uniformly, its temperature is given by
$\theta ={20}^{\circ }C+\left(\frac{{80}^{\circ }C-{20}^{\circ }C}{300s}\right)t={20}^{\circ }C+\left(\frac{1^{\circ }C}{5s}\right)t$,
where t = time
$\therefore \frac{dQ}{dt}=k(\theta -{\theta }_0)$
$=\left(\frac{1}{2}W{/}^{\circ }C\right)\left(\frac{1^{\circ }C}{5s}\right)t=\left(\frac{1}{10}W/s\right)t.$
$\therefore Q={\int }_{0s}^{300s}\left(\frac{1}{10}W/s\right)tdt$
$=\left(\frac{1}{10}W/s\right)\frac{(300s{)}^2-(0s{)}^2}{2}$
$=\left(\frac{1}{20}W/s\right)\left(90000s^2\right)$
$=4500Ws=4500J$