Question

$3.75\times {10}^5$ calories of heat is given out by $5kg$ of water at $100°C$. Calculate the temperature of the cooled water. The specific heat capacity of water is $1000calK{g}^{-1}°C^{-1}$.

Answer 1

Step 1: Information

Given

mass$\left(m\right)=5kg$

Heat supplied$=3.75\times {10}^{5}cal$

Final temperature (T)=$100°C$

Find:

The temperature of the cooled water

Step 2: Calculating the temperature of cooled water

Heat supplied is given as $mc{\theta }_R$

$$\begin{gathered} 3.75\times {10}^5=5\times 1000\times {\theta }_R \\ {\theta }_R=\frac{3.75\times {10}^5}{5000}=\frac{375}{5} \\ {\theta }_R=75°C \end{gathered}$$

Rise in temperature = Finial temperature-initial temperature

$$\begin{gathered} 75=100-x \\ x=100-75 \\ x=25°C \end{gathered}$$

Hence temperature of cooled water is $25°C$