Question

$400g$ of mercury of specific heat capacity $0.14J{g}^{-1}°C^{-1}$ is heated by a $200W$ heater for $1$ min and $40s$. If initially, mercury is at $0°C$, calculate its final temperature.

Answer 1

Step 1

Given:

1. Power of heater (P) $200W$,
2. Mass of mercury(m) $400g$
3. Time(t) $1min40s=100s$

Find:

The final temperature of mercury

Step 2:

Let the final temperature is x

Change in temperature is ${\theta }_R=x-0°C$

The heat energy supplied by the heater is given as $P\times t=200\times 100=20000J$

Step 3:

By the law of the conservation of energy we have

$$\begin{gathered} mc{\theta }_R=20000 \\ 400\times 0.14\times (x-0)=20000 \\ x=\frac{20000}{400\times 0.14} \\ x=357.1°C \end{gathered}$$

Hence, the final temperature of the mercury will be $357.1°C$