Question

A vessel contains $110\text{g}$ of water. The heat capacity of the vessel is equal to that of $10\text{g}$ of water. The initial temperature of water in the vessel is ${10}^{\circ }\text{C}$. While the water and vessel are in thermal equilibrium, $220\text{g}$ of hot water at ${70}^{\circ }\text{C}$ is poured into the vessel. Find the final temperature of the vessel + water system (neglecting radiation loss).

• A. ${75}^{\circ }\text{C}$
• B. ${81}^{\circ }\text{C}$
• C. ${63}^{\circ }\text{C}$
• D. ${49}^{\circ }\text{C}$

Answer 1

The correct option is D ${49}^{\circ }\text{C}$

Given,
Mass of water in the vessel $\left(m_1\right)=110\text{g}$
Initial temperature of water $\left({\theta }_1\right)={10}^{\circ }\text{C}$
Initial temperature of hot water $\left({\theta }_2\right)={70}^{\circ }\text{C}$
Mass of hot water added to the vessel$\left(m_2\right)=220\text{g}$
Also that, the heat capacity of vessel is equal to that of $10\text{g}$ of water.
$\Rightarrow$ Heat capacity of vessel $=m_v{c}_v=10\times 1=10\text{cal}/{}^{\circ }\text{C}$
Let the final temperature of the water + vessel system be $\theta$.
From the principle of calorimetry,
Heat taken by water + vessel system = Heat given by hot water
$\Rightarrow m_1{c}_w(\theta -{\theta }_1)+m_v{c}_v(\theta -{\theta }_1)=m_2{c}_w({\theta }_2-\theta )$
$\Rightarrow 110\times 1(\theta -10)+10(\theta -10)=220\times 1(70-\theta )$
$\Rightarrow \theta ={48.8}^{\circ }\text{C}\approx {49}^{\circ }\text{C}$
Thus, option (d) is the correct answer.