Question

When $M_1$ grams of ice at $-10{}^{\circ }\text{C}$ $(\text{Specific heat}=0.5{\text{cal g}}^{-1}{}^{\circ }{\text{C}}^{-1})$ is added to $M_2$ grams of water at ${50}^{\circ }\text{C}$, it is observed that no ice is left and the water is at $0^{\circ }\text{C}$. Which of the following correctly gives the value of latent heat of ice, (in ${\text{cal g}}^{-1}$) is :

• A. $\frac{50M_2}{M_1}-5$
• B. $\frac{50M_2}{M_1}$
• C. $\frac{50M_1}{M_2}-5$
• D. $\frac{5M_1}{M_2}-50$

Answer 1

The correct option is A $\frac{50M_2}{M_1}-5$

Using the principle of calorimetry,

The heat gained by the ice to completely melt is equal to the heat lost by the water to cool down to $0{}^{\circ }\text{C}$

$M_1\times C_{ice}\times \left(10\right)+M_{1}L=M_2\times C_{\text{w}}\times \left(50\right)$

$M_1\times \left(0.5\right)\times 10+M_{1}L=M_2\times 1\times 50$

$\Rightarrow L=\frac{50M_2}{M_1}-5$

Hence, option $\left(A\right)$ is correct.