Question

A metal ball of mass $1\text{kg}$ is heated by means of a $20\text{W}$ heater in a room at $20{}^{\circ }\text{C}$. The temperature of the ball becomes steady at $50{}^{\circ }\text{C}$. Assuming Newton's law of cooling and that temperature of the ball rises uniformly from $20{}^{\circ }\text{C}$ to $30{}^{\circ }\text{C}$ in $5\text{minutes}$, find the specific heat capacity of the metal.

• A. $700\text{J}{\text{kg}}^{-1}$${\text{K}}^{-1}$
• B. $600\text{J}{\text{kg}}^{-1}$${\text{K}}^{-1}$
• C. $500\text{J}{\text{kg}}^{-1}$${\text{K}}^{-1}$
• D. $400\text{J}{\text{kg}}^{-1}$${\text{K}}^{-1}$

Answer 1

The correct option is C $500\text{J}{\text{kg}}^{-1}$${\text{K}}^{-1}$

When the temperature of the ball become steady at $50{}^{\circ }\text{C}$, the maximum power radiated by the ball will be equal to the power supplied by the heater.

Thus, $P_0=20\text{W}$

At any temperature $T$, power $P$, supplied by the heater,

$P=P_0\left(\frac{T-T_0}{T_{max}-T_0}\right)$

At $30{}^{\circ }\text{C}$, power supplied by the heater,
$P_1=20\left(\frac{{30}^{\circ }-{20}^{\circ }}{{50}^{\circ }-{20}^{\circ }}\right)=\frac{20}{3}\text{W}$

At $20{}^{\circ }\text{C}$, power supplied by the heater,
$P_2=20\left(\frac{{20}^{\circ }-{20}^{\circ }}{{50}^{\circ }-{20}^{\circ }}\right)=0\text{W}$

Since the change in temperature is uniform so, average power supplied by the heater,$\therefore P_{average}=\frac{P_1+P_2}{2}=\frac{20/3+0}{2}=\frac{10}{3}\text{W}$

So, the energy radiated in $5\text{minutes}$

$E_r=P_{average}\times \text{time}$

$E_r=\frac{10}{3}\times 5\times 60=1000\text{J}$

Energy gained by ball in $5\text{minutes}$,

$E_g=P_0\times \text{time of supplying energy}$

$E_g=20\times 5\times 60=6000\text{J}$

$\therefore$ Net gained energy by ball,
$\Delta E=E_g-E_r=6000-1000$

$\therefore \Delta E=5000\text{J}$

We know that,
$\Delta E=mS\Delta T..........\left(1\right)$

Where, mass of the ball, $m=1\text{kg}$ and $S$ is a specific heat of the ball.
Substituting the value in equation $\left(1\right)$

$\Rightarrow 5000=1\times S\times ({30}^{\circ }-{20}^{\circ })$

$\therefore S=500\text{J}{\text{kg}}^{-1}$${\text{K}}^{-1}$