You have 15 grams of ice at 0°C. How much energy is needed to melt the ice and warm it to 25°C. Ice melts at 0°C. The specific heat of water is 4.18 J/g*C. The heat of fusion of water is 80. calories per gram O a. 0.0075 calories O b. 380 calories O c. 1200 calories O d. 1600 calories Oe. 30,000 calories

The heat of fusion of water, L, is the amount of heat per gram required to melt the ice to water, a process which takes place at a constant temperature of 0 °C. The specific heat of water, c, is the amount of heat required to change the temperature of 1 gram of water by 1 degree Celsius.

We will convert the units of c from Jg⁻¹°C⁻¹ to cal·g⁻¹°C⁻¹ since the answers are provided in calories. The conversion factor is 4.18 J/cal.

(4.18 Jg⁻¹°C⁻¹)(cal/4.18J) = 1 cal·g⁻¹°C⁻¹

First we calculate the heat required to melt the ice, where M is the mass:

Q = ML = (15 g)(80 cal/g) = 1200 cal

Then, we calculate the heat required to raise the temperature of water from 0 °C to 25 °C.

Q = mcΔt = (15 g)(1 cal·g⁻¹°C⁻¹)(25 °C - 0 °C) = 380 cal

The answer is rounded so that there are two significant figures

The total heat required for this process is (1200 cal + 380 cal) = 1580 cal

The rounded answer is 1600 calories.