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: $Q_2=mL$ $=1\times 80=80calorie$ Heat required to boil the water from $0{}^{\circ }C$ to $100{}^{\circ }C$ $Q_3=ms\Delta t$ $=1\times 1\times 100$ $=100calorie$ Heat required to change 1 g of water to vapour is $Q_4=mL$ $=1\times 540$ $=540calorie$ $\therefore$ Total heat required in whole process $Q=Q_1+Q_2+Q_3+Q_4$ $=25+80+100+540$ $=745calorie$