When M 1 gram of ice at - 10 o C (specific heat = 0.5   c a l   g - 1   ℃ - 1 ) is…

When M1 gram of ice at -10oC (specific heat =0.5 cal g-1 ℃-1 ) is added to M2 gram of water at 50 oC, finally no ice is left and the water is at 0 oC . The value of latent heat of ice, in cal g-1 is:
  1. 50M2M1
  2. 5M1M2-50
  3. 5M2M1-5
  4. 50M2M1-5

Solution

Principle of calorimetry
M1Sice0--10+M1Lf=M2Swater50-0
M10.510+M1Lf=M250
5M1+M1Lf=M250
Lf=50M2M1-5

Asked in: JEE Main 2019 (12 Apr Shift 1)

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