Question

To treat a burn on your hand, you decide to place an ice cube on the burned skin. The mass of the ice cube is 12.8 g, and its initial temperature is -12.5 °C. The water resulting from the melted ice reaches the temperature of your skin, 30.9 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand. Constants may be found here.

Answer 1

mass of the ice cube = 12.8 g

Initial temperature T1 = - 12.5 °C

Final temperature T2 = 30.9 °C

Heat absorbed by raising the temperature of ice from - 12.5°C to 0°C

Q1 = m x Cp x [0 - (-12.5)]

= 12.8 g x 2.108 J/g-K x (12.5K)

= 337.28 J

Phase change at 0°C

Q2 = m x Hfus

= 12.8 g x 334 J/g

= 4275.2 J

Heat absorbed by raising the temperature of water from 0°C to 30.9°C

Q3 = m x Cp x [30.9 - 0]

= 12.8 g x 4.184 J/g-K x (30.9)K

= 1654.86 J

Total heat absorbed = Q1 + Q2 + Q3

= 337.28 + 4275.2 + 1654.86

= 6267.34 J