Question

When $M_1$ gram of ice at $-10^{\circ}C$ (specific heat = $0.5 \; cal \; g^{-1 \circ } C^{-1}$) is added to $M_2$ gram of water at $50^{\circ}C$, finally no ice is left and the water is at $0^{\circ}C$. The value of latent heat of ice, in cal $g^{-1}$ is:

• A. $\frac{5M_{1}}{M_{2}} - 50$
• B. $\frac{50M_{2}}{M_{1}}$
• C. $\frac{50M_{2}}{M_{1}} - 5$
• D. $\frac{5M_{2}}{M_{1}} - 5$

Answer 1

Answer: (C)

$\frac{50M_{2}}{M_{1}} - 5$

Answer 2

$Heat\, lost = Heat \,gain$
$\Rightarrow \; M_2 \times 1 \times 50 = M_1 \times 0.5 \times 10 + M_1.L_f$
$\Rightarrow \; L_f = \frac{50 M_2 - 5 M_1}{M_1}$
$= \frac{50 M_2 }{M_1} - 5$