Question:

When M1 gram of ice at -100°C (specific heat = 0.5 cal/g°C) 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, is:

5M1M2-50

5M2M1

50M2M1-5

5M2M1-5

Correct option is C. 50M2M1-5
Heat lost = Heat gain
⇒M2 × 1 × 50 = M1 × 0.5 × 10 + M1.Lf
⇒Lf = 50M2 - 5M1
M1 = 50M2M1 - 5