Properties Of Solids And Liquids Question 29

Question: Work done in converting one gram of ice at -10°C into steam at 100°C is [MP PET/PMT 1988; EAMCET (Med.) 1995; MP PMT 2003]

Options:

A) 3045 J

B) 6056 J

C) 721 J

D) 616 J

Show Answer

Answer:

Correct Answer: A

Solution:

Ice (-10°C) converts into steam as follows (ci = Specific heat of ice, cW = Specific heat of water)

Total heat required $ Q=Q _{1}+Q _{2}+Q _{3}+Q _{4} $

therefore $ Q=1\times 0.5(10)+1\times 80+1\times 1\times (100-0)+1\times 540 $

$ =725cal $

Hence work done $ W=JQ=4.2\times 725=3045J $



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