The specific heat of ice is 0.48 cal/g-oC. The heat of fusion of ice is 80 cal/g. What will?
A)0C
B)298 K
C)100oC
D)104oC
Answer:
Use the formula C = Q/m x deltaT
C = specific heat capacity
Q = heat absorbed
m= mass of object
deltaT = change in temp.
0.48cal/gC = 1000Cal/20g x delta T
deltaT = 104.2C. Now you add the temperature difference from -10C. Since you add energy to ice, the ice can only melt even more(not increase temp.) The question is not telling you that the final temp. should be zero C because there
can be more energy added that will bring up the temperature.
To take 20 g of ice at -10 to 0 deg C will require 20g x 10 deg x 0.48 calories per g/deg C = 96 calories. You have 1000 - 96 = 904 calories left to melt the ice. 904 calories / 80 calories per gram = 11.3 grams of ice will melt. The final result is 11.3 g ice and 8.7 g water all at 0 deg C. Answer A.
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