In order to melt $\underline { \text { two } }$ grams of ice, how many calories of heat energy must be added to the ice during its phase change? (specific heat of water = 1 cal/gm; heat of fusion = 85 cal/gm)
(heat of vaporization = 585 cal/gm; heat of sublimation = 670 cal/gm)
A) 2
B) 170
C) 172
D) 1170
E) 1340

Question

In order to melt $\underline { \text { two }  }$ grams of ice, how many calories of heat energy must be added to the ice during its phase change? (specific heat of water = 1 cal/gm; heat of fusion = 85 cal/gm)
(heat of vaporization = 585 cal/gm; heat of sublimation = 670 cal/gm)
A) 2
B) 170
C) 172
D) 1170
E) 1340

Quizplus · Accepted Answer

To melt ice, the heat of fusion is used. For 2 grams of ice, the energy required is 2 grams * 85 cal/gm = 170 calories.

In Order to Melt two ‾\underline { \text { two } } two ​

In Order to Melt $\underline { \text { two } }$