Answer:
d. 1600 calories
Explanation:
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.
