Properties Of Solids And Liquids Question 25
Question: Heat required to convert one gram of ice at 0°C into steam at 100°C is (given Lsteam = 536 cal/gm) [Pb. PMT 1990]
Options:
A) 100 calorie
B) 0.01 kilocalorie
C) 716 calorie
D) 1 kilocalorie
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Answer:
Correct Answer: C
Solution:
Conversion of ice (0°C) into steam (100°C) is as follows Heat required in the given process $ =Q _{1}+Q _{2}+Q _{3} $
$ =1\times 80+1\times 1\times (100-0)+1\times 536=716cal $
BETA
