When M1 gram of ice at -10 °C (specific heat=0.5cal g-1 °C -1) is added to M2 gram of water at 50 °C , finally no ice is left and the water is at 0°C . The value of latent of ice, in cal g-1 is :

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When M₁gram of ice at -10 °C (specific heat=0.5cal g^-1 °C ^-1) is added to M₂ gram of water at 50 °C , finally no ice is left and the water is at 0°C . The value of latent of ice, in cal g^-1 is :

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

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

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

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

answer is A.

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Detailed Solution

$\begin{array}{l} M_{1} C_{ice} \times (10) + M_{1} L = M_{2} C_{ω} (50) \\ Or M_{1} C_{ice} \times (= 0 .5) \times 10 + M_{1} L = M_{2} \times 1 \times 50 \\ \Rightarrow L = \frac{50 M_{2}}{M_{1}} - 5 \end{array}$