PhysicsEasy251×since 2002Q7453When 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 :A50M2M1−5{{50{M_2}} \over {{M_1}}} - 5M150M2−5B50M2M1{{50{M_2}} \over {{M_1}}}M150M2C5M2M1−5{{5{M_2}} \over {{M_1}}} - 5M15M2−5D5M1M2−50{{5{M_1}} \over {{M_2}}} - 50M25M1−50Check answerSkip