Final answer:
To calculate the heat absorbed when 185 g of ice at 0.0 °C melts and warms to 37 °C, you must add the heat of fusion for melting the ice, which is 14.8 kcal, to the heat needed to raise the temperature of the water to 37 °C, which is 6.845 kcal, for a total of 21.645 kcal.
Step-by-step explanation:
The question is asking to calculate the amount of heat in kcal absorbed when 185 grams of ice at 0.0 °C melts and then is raised to the body temperature of 37.0 °C. To find this, we need to consider both the heat of fusion to melt the ice and the specific heat capacity of water to increase its temperature.
First, we use the heat of fusion for ice, which is 80 cal/g, to find how much heat is required to melt 185 g of ice:
(185 g) × (80 cal/g) = 14800 cal
Now, convert calories to kcal:
14800 cal ÷ 1000 cal/kcal = 14.8 kcal
This is the heat absorbed to just melt the ice. Next, we need to calculate the heat absorbed to raise the temperature from 0.0 °C to 37 °C:
The specific heat capacity of water is 1 cal/g°C, so the heat required is:
(185 g) × (1 cal/g°C) × (37 °C) = 6845 cal
Convert to kcal:
6845 cal ÷ 1000 cal/kcal = 6.845 kcal
Then add both parts:
14.8 kcal (melting) + 6.845 kcal (raising temperature) = 21.645 kcal
So the total heat absorbed by the ice and then water is 21.645 kcal.