Question

$100\text{g}$ of ice $\left(\text{latent heat}80{\text{cal g}}^{-1}\right)$ at $0^{\circ }\text{C}$ is mixed with $100\text{g}$ of water (specific heat $1{\text{cal g}}^{-1\circ }{\text{C}}^{-1}$) at ${80}^{\circ }\text{C}$. The final temperature of the mixture will be

• A. $0^{\circ }\text{C}$
• B. ${40}^{\circ }\text{C}$
• C. ${80}^{\circ }\text{C}$
• D. $<0^{\circ }\text{C}$

Answer 1

The correct option is A $0^{\circ }\text{C}$

Given that,
Latent heat of fusion of ice $\left(L_f\right)=80\text{cal/g}$
Specific heat capacity of water $\left(c_w\right)=1{\text{cal/g}}^{\circ }\text{C}$
Mass of ice $\left(m_i\right)=100\text{g}$
Mass of water $\left(m_w\right)=100\text{g}$
Initial temperature of ice $\left(T_1\right)=0^{\circ }\text{C}$
Initial temperature of water $\left(T_2\right)={80}^{\circ }\text{C}$
Let the temperature of mixture be $T$.

Heat required to melt $100\text{g}$ of ice to water is $Q_1=m_i{L}_f=100\times 80=8000\text{cal}$
& Heat required to reduce the temperature of water from ${80}^{\circ }\text{C}$ to $0^{\circ }\text{C}$ is
$Q_2=m_w{c}_w\Delta T=100\times 1\times (80-0)=8000\text{cal}$
$\because Q_1=Q_2$
From the priciple of calorimetry, we can say that heat released from the cooling of water at $80{}^{\circ }\text{C}$ will used to just melt the given mass of ice. At equilibrium, temperature of the system will be $0^{\circ }\text{C}$.
Thus, option (a) is the correct answer.