Question

An ice at -50?C is to be converted into vapour at 100?C. The heat required in this process is : (specific heat of ice = 0.5 calorie/gram 0C Latent heat of ice = 80 calorie/gram Latent heat of vapour =540 calorie/gram)

• A. 1028 ealorie
• B. 624 calorie
• C. 745 calorie
• D. 810 calorie

Answer 1

Answer: (C)

Heat required to raise the temperature of ice from $ -50{}^\circ C $ to $ 0{}^\circ C $ is $ {{Q}_{1}}=ms\Delta t=1\times 0.5\times 50 $ =25 calorie Heat required to liquify 1 g ice at $ 0{}^\circ C $ to g water at $ 0{}^\circ C $ is: $ {{Q}_{2}}=mL $ $ =1\times 80=80\text{ }calorie $ Heat required to boil the water from $ 0{}^\circ C $ to $ 100{}^\circ C $ $ {{Q}_{3}}=ms\Delta t $ $ =1\times 1\times 100 $ $ =100\text{ }calorie $ Heat required to change 1 g of water to vapour is $ {{Q}_{4}}=mL $ $ =1\times 540 $ $ =540\text{ }calorie $ $ \therefore $ Total heat required in whole process $ Q={{Q}_{1}}+{{Q}_{2}}+{{Q}_{3}}+{{Q}_{4}} $ $ =25+80+100+540 $ $ =745\text{ }calorie $