Question

A sports trainer applies an ice bag to the back of an injured athlete. Calculate the heat in kcal that is absorbed if 165g of ice at 0.0∘C is placed in an ice bag, melts, and rises to body temperature of 37.0∘C. (For water, 80. cal (334 J) is needed to melt 1g of ice or must be removed to freeze 1g of water.)

Express the heat to two significant figures and include the appropriate units.

Answer 1

The heat absorbed by the ice is given by the formula: heat = mass * specific heat capacity * change in temperature.

The mass of the ice is 165g. The specific heat capacity of water is 4.18 J/g°C. The change in temperature is 37.0°C - 0.0°C = 37.0°C.

The heat absorbed by the ice is: 165g * 4.18 J/g°C * 37.0°C = 29,894.5 J

To convert from J to kcal: 1 kcal = 4184 J. Therefore,

29,894.5 J / 4184 J/kcal = 7.13 kcal

The heat absorbed by the ice is 7.13 kcal