Principle of calorimetry
M1Sice(0−(−10))+M1Lf=M2Swater(50−0)
M1(0.5)(10)+M1Lf=M2(50)
5M1+M1Lf=M2(50)
Lf=M150M2−5
JEE Main 2019 — Physics Thermodynamics
When M1 gram of ice at −10oC (specific heat =0.5calg−1℃−1 ) is added to M2 gram of water at 50oC, finally no ice is left and the water is at 0oC . The value of latent heat of ice, in calg−1 is:
Held on 12 Apr 2019 · Verified 6 Jul 2026.
M150M2
M25M1−50
M15M2−5
M150M2−5
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