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

Engineering

Physics

Calorimetry

Question

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

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

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

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

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

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Heat lost = Heat gain

⇒ M₂ × 1 × 50 = M₁ × 0.5 × 10 + M₁,L_f

$\Rightarrow L_{f} = \frac{50 M_{2} - 5 M_{1}}{M_{1}}$

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