JEE Main 2019 — Physics solution

JEE Main2019Physics

When M₁ gram of ice at –10°C (specific heat = 0.5 cal 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 heat of ice, in cal g^–1 is :

A
$\frac{50 M _{2}}{M _{1}} - 5$
B
$\frac{5 M _{1}}{M _{2}} - 50$
C
$\frac{50 M _{2}}{M _{1}}$
D
$\frac{5 M _{2}}{M _{1}} - 5$

Solution

$M_{1} \times 5 + M_{1} L = M_{2} 50 L = \frac{50 M _{2}}{M _{1}} - 5$

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