Ten grams of ice at -20°C is to be changed to steam at 130°C.The specific heat of both ice and steam is 0.5 cal/g . °C.The specific heat of water is 1.00 cal/g . K.The heat of fusion is 80 cal/g and the heat of vaporization is 540 cal/g.The entire process requires:

Question

Quizplus · Accepted Answer

To find the total heat required, we need to calculate the heat required to:
1. Raise the temperature of ice from -20F1F1F1S1F1F1F10C to 0F1F1F1S1F1F1F10C
2. Melt the ice at 0F1F1F1S1F1F1F10C
3. Raise the temperature of water from 0F1F1F1S1F1F1F10C to 100F1F1F1S1F1F1F10C
4. Vaporize the water at 100F1F1F1S1F1F1F10C
5. Raise the temperature of steam from 100F1F1F1S1F1F1F10C to 130F1F1F1S1F1F1F10C

1. Q = m*C*ΔT = 10*0.5*(0-(-20)) = 100 cal
2. Q = m*Hf = 10*80 = 800 cal
3. Q = m*C*ΔT = 10*1*(100-0) = 1000 cal
4. Q = m*Hv = 10*540 = 5400 cal
5. Q = m*C*ΔT = 10*0.5*(130-100) = 150 cal

Total heat required = 100+800+1000+5400+150 = 7450 cal

Therefore, the correct answer is D) 7450 cal.

Ten Grams of Ice at -20°C Is to Be Changed