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 17.4 g, and its initial temperature is -10.7 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.5 °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 block(m) = 17.4g

Total Heat absorbed = Heat absorbed when Temp of ice changes from -10.7C to 0C + Latent heat of fusion of ice to water(at 0C) + Heat absorbed when Temp of ice changes from 0C to 28.5C

Heat of fusion of ice (l_f at 0C) = 334J/gram

C_p,liquid(specific heat capacity) of water = 4.184 J/gram . C

C_p,solid(specific heat capacity) of ice = 2.108 J/gram . C

So,

Total Heat absorbed = m *C_p,solid * 10.7 + m * l_f + m * C_p,liquid * 28.5

= 17.4 * ( 2.108 *10.7 + 334 + 28.5*4.184) J

= 8.279 kJ