Question

3.75 x 10⁵ calories of heat is given out by 5 kg of water at 100 °C. Calculate the temperature of cooled water. The specific heat capacity of water is 1000 cal kg^-1 °C^-1.

Answer 1

Given:

The heat given out, $H=3.75\times {10}^{5}calories$

The mass of the water, $m=5kg$

The initial temperature of water $=100°C$

The specific heat capacity of water, $c=1000calk{g}^{-1}°C^{-1}$

Finding temperature of cooled water

Let the final temperature of the water$=x$

We know that, the heat given out by a body, $H=mc{\theta }_f$ where $m$ is the mass of the body, ${\theta }_f$ is the fall in temperature of the body and $c$ is the specific heat capacity.

By substituting the given values in the above expression, we get

$$\begin{gathered} 3.75\times {10}^5=5\times 1000\times (100-x) \\ \Rightarrow 3.75\times {10}^5=5000\times (100-x) \\ \Rightarrow 3.75\times {10}^5=500,000-5,000x \\ \Rightarrow 3.75\times {10}^5=5000(100-x) \\ \Rightarrow \frac{3.75\times {10}^5}{5\times {10}^3}=100-x \\ \Rightarrow 0.75\times {10}^2=100-x \\ \Rightarrow 75=100-x \\ \Rightarrow -25=-x \\ \Rightarrow x=25 \end{gathered}$$

Therefore, the temperature of cooled water is 25 °C.