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

Question

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

${{50{M_2}} \over {{M_1}}} - 5$

${{50{M_2}} \over {{M_1}}}$

${{5{M_2}} \over {{M_1}}} - 5$

${{5{M_1}} \over {{M_2}}} - 50$

Solution

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