Ten grams of ice at -20$\degree$C is to be changed to steam at 130$\degree$C. The specific heat of both ice and steam is 0.5 cal/g .$\degree$C. 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

First, we need to determine how much heat is required to take the ice from -20°F to 0°F:
Q = m*c*ΔT = 10g * 0.5 cal/g°F * (0°F - (-20°F)) = 100 cal

Next, we need to determine how much heat is required to melt the ice:
Q = m*L_f = 10g * 80 cal/g = 800 cal

Then, we need to determine how much heat is required to take the water (now at 0°F) to 100°C:
Q = m*c*ΔT = 10g * 0.5 cal/g°F * (100°C - 0°F) = 500 cal

After that, we need to determine how much heat is required to vaporize the water:
Q = m*L_v = 10g * 540 cal/g = 5400 cal

Finally, we need to determine how much heat is required to take the steam from 100°C to 130°F:
Q = m*c*ΔT = 10g * 0.5 cal/g°F * (130°F - 100°C) = 250 cal

The total heat required is the sum of all of these values:
100 + 800 + 500 + 5400 + 250 = 7450 cal
Therefore, the answer is D) 7450 cal.

Ten Grams of Ice at -20 °\degree° C Is to Be Changed to Steam at 130

Ten Grams of Ice at -20 $\degree$ C Is to Be Changed to Steam at 130