Question

A metal ball of mass $1.0\text{kg}$ is kept in a room at ${15}^{\circ }\text{C}$. It is heated using a heater. The heater supplies heat to the ball at a constant rate of $24\text{W}$. The temperature of the ball rises as shown in the graph. Assuming that Newton's law of cooling can be used, calculate the rate of heat loss from the ball when it is at a temperature of ${20}^{\circ }\text{C}$.

• A. $2.2\text{W}$
• B. $4\text{W}$
• C. $6.3\text{W}$
• D. $1\text{W}$

Answer 1

The correct option is B $4\text{W}$

Given, mass of the metal ball $\left(m\right)=1.0\text{kg}$
At steady state, no heat is absorbed or lost [no change in temperature].
i.e Rate of loss of heat at steady state is equal to the power supplied.
$\Rightarrow H=\frac{dQ}{dt}=24\text{W}$

From Newton's law of cooling, we know that
Average rate of cooling at steady state $=-k\times$ Temperature difference between the body and surroundings.
$-\frac{dQ}{dt}=k[T_1-T_s]$
Here, $T_1={45}^{\circ }\text{C},T_s={15}^{\circ }\text{C}$
$\Rightarrow -24=k\times (45-15)$
$\Rightarrow k=-0.8{\text{W/}}^{\circ }\text{C}......\left(1\right)$
Thus, rate of heat loss at ${20}^{\circ }\text{C}$
$P=k\times (20-15)......\left(2\right)$
Using $\left(1\right)$ in $\left(2\right)$, we get
$P=-0.8\times 5=4\text{W}$
Thus, option (b) is the correct answer.