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When $M_{1}$ gram of ice at $- 10^{0} C$ (Specific heat $= 0.5 c a l g^{- 1}^{0} C^{- 1}$ ) is added to $M_{2}$ gram of water at $50^{0} C$ , finally no ice is left and the water is at $0^{0} C$ . The value of latent heat of ice, in $c a l g^{- 1}$ is:

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

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

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

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

answer is A.

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Detailed Solution

$M_{1} C_{i c e} \times (10) + M_{1} L = M_{2} C_{ω} (50) \Rightarrow M_{1} \times C_{i c e} (= 0.5) \times 10 + M_{1} L = M_{2} \times 1 \times 50 \Rightarrow L = \frac{50 M_{2}}{M_{1}} - 5$

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