Question

Water's heat of fusion is 80. cal/g

, its specific heat is 1.0 calg⋅∘C
, and its heat of vaporization is 540 cal/g
.
A canister is filled with 390 g
of ice and 100. g
of liquid water, both at 0 ∘C
. The canister is placed in an oven until all the H2O
has boiled off and the canister is empty. How much energy in calories was absorbed?

Express your answer to two significant figures and include the appropriate units.

Answer 1

The total energy absorbed by the canister containing 390 g of ice and 100 g of water in the oven is 344,800 cal. This includes converting ice to water, heating the water to boiling, and then vaporizing the water into steam.

Step-by-step explanation:

To find out the total energy absorbed by the canister of ice and water in the oven, we calculate the energy required for each phase change and temperature change the water undergoes. We have 390 g of ice and 100 g of liquid water, all starting at 0 °C.

• Firstly, we need to convert all the ice to water. The heat of fusion for water is 80 cal/g. So, for 390 g of ice, the heat required is 390 g × 80 cal/g = 31,200 cal.
• Next, we raise the temperature of the now 490 g of water (390 g from the melted ice and 100 g initial water) from 0 °C to 100 °C. The specific heat of water is 1.0 cal/g°C, therefore the heat required is 490 g × 100 °C × 1.0 cal/g°C = 49,000 cal.
• Finally, we convert the liquid water to steam. The heat of vaporization for water is 540 cal/g. So for 490 g of water, the heat required is 490 g × 540 cal/g = 264,600 cal.

Add these three quantities to get the total heat absorbed: 31,200 cal + 49,000 cal + 264,600 cal = 344,800 cal.

Therefore, the total energy absorbed by the canister in calories is 344,800 cal.