Physics

Physics

Problem

When M1M_1 gram of ice at −10∘C-10^\circ\text{C} (Specific heat = 0.5 cal g−1 ∘C−10.5 \text{ cal g}^{-1} \text{ }^\circ\text{C}^{-1} ) is added to M2M_2 gram of water at 50∘C50^\circ\text{C} , finally no ice is left and the water is at 0∘C0^\circ\text{C} . The value of latent heat of ice, in cal g−1\text{cal g}^{-1} is: a. 50M2M1−5\frac{50M_2}{M_1} - 5 b. 5M1M2−50\frac{5M_1}{M_2} - 50 c. 50M2M1\frac{50M_2}{M_1} d. 5M2M1−5\frac{5M_2}{M_1} - 5

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