Question

A sports trainer applies an ice bag to an athlete’s shoulder. Calculate the heat absorbed by the ice bag if there are 140 grams of ice at 0.0 °C and raises to body temperature 37.0 °C. Give your answer in kilocalories (kcal).

Answer 1

The ice bag absorbs a total of 16.38 kilocalories of heat as it melts and then raises the temperature of the resulting water from 0.0 °C to 37.0 °C, with an initial mass of 140 grams.

Step-by-step explanation:

To calculate the heat absorbed by the ice bag as it goes from 0.0 °C to 37.0 °C, we need to account for two processes: the melting of ice and the heating of water.

First, we calculate the heat required to melt the ice using the heat of fusion for water, which is 80 cal/g. Then, we calculate the heat needed to raise the temperature of the resulting water from 0.0 °C to 37.0 °C using the specific heat capacity for water, which is 1 cal/(g°C).

• Heat needed for melting ice: Q1 = mass * heat of fusion = 140 g * 80 cal/g
• Heat needed for heating water: Q2 = mass * specific heat capacity * ΔT (change in temperature) = 140 g * 1 cal/(g°C) * (37.0 °C - 0.0 °C)

Accordingly, Q1 = 11200 calories and Q2 = 5180 calories. Now, we convert these values to kilocalories.

Q1 = 11.2 kcal and Q2 = 5.18 kcal

Finally, we sum the calories for both processes to find the total heat absorbed by the ice bag.

Total heat absorbed: 16.38 kcal

The ice bag absorbs 16.38 kilocalories of heat energy when raising its temperature from 0.0 °C to 37.0 °C with an initial mass of 140 grams.