Question

Certain amount of hot water was added to three times its mass of cold water at ${10}^{\circ }C$. The resulting temperature was found to be ${20}^{\circ }C$. The initial temperature of the hot water is:

• A. ${20}^{\circ }C$
• B. ${50}^{\circ }C$
• C. ${30}^{\circ }C$
• D. ${10}^{\circ }C$

Answer 1

The correct option is B ${50}^{\circ }C$

According to the given situation, the heat is transferred from hot water to cold water and temperature of hot water falls with a rise in temperature of cold water subsequently.
To calculate heat transfer,
Let the mass of water be $m$, let the initial temperature of hot water be $t^{\circ }C$ and $c$ be the specific heat capacity.
Initial temperature of cold water, $t_1={10}^{\circ }C$ and the final temperature after mixing, $t_2={20}^{\circ }C$
Heat transfer $=$(mass)$\times$(specific heat)$\times$(temperature change)
or, $H=m\times c\times \Delta T$
Then according to the given condition, we have,
$m\times c\times (t-t_1)=3m\times c\times (t_2-t_1)$
$\Rightarrow m\times c\times (t-20)=3m\times c\times (20-10)$
or, $t-20=3(20-10)$
or, $t=30+20={50}^{\circ }C$

Hence, the temperature of hot water was ${50}^{\circ }C$.
This implies that on mixing hot and cold water, temperature of hot water drop to ${20}^{\circ }C$ from ${50}^{\circ }C$ and temperature of cold water rises from ${10}^{\circ }C$ to ${20}^{\circ }C$.