Heat and Thermodynamics 1 Question 1

1. When M1 gram of ice at 10C (specific heat =0.5cal g1C1 ) is added to M2 gram of water at 50C, finally no ice is left and the water is at 0C. The value of latent heat of ice, in cal g1 is

(a) 50M2M15

(b) 50M1M250

(c) 50M2M1

(d) 5M2M15

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Solution:

  1. Key Idea In such kind of heat transfer problems, Heat given by water = Heat gained by ice and (heat) )liquid = mass × specific heat × (Heat) )solid =( mass × latent heat ) temperature =mlslΔTl
  • (mass × specific heat × temperature) =(ms×L)+ms×ss×ΔTs

Heat given by water is (specific heat of water is 1calg1C1 )

(ΔH)water =M2×1×(500)=50M2

Heat taken by ice is

(ΔH)ice =M1×0.5×[0(10)]+M1×Lice (ΔH)ice =5M1+M1Lice 

Comparing Eqs. (i) and (ii), we get

(ΔH)water =(ΔH)ice 50M2=5M1+M1Lice Lice =50M25M1M1Lice =50M2M15



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