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 $
