question_answer

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]

A)            100 calorie                             

B)            0.01 kilocalorie

C)            716 calorie                             

D)            1 kilocalorie

Correct Answer: C

Solution :

                   Conversion of ice (0°C) into steam (100°C) is as follows                    Heat required in the given process \[={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}\]            \[=1\times 80+1\times 1\times (100-0)+1\times 536=716\,cal\]