To treat a burn on your hand you decide to place an ice cube

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 15.9 g, and its initial temperature is -10.3 degree C. The water resulting from the melted ice reaches the temperature of your skin, 28.2 degree C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand. Constant may be found here.

Solution

Here ,

mass of cube , m = 15.9 gm

m = 0.0159 Kg

T1 = -10.3 degree C

T2 = 28.2 degree C

specific heat of water , Sw = 4186 J/(kg/degree C)

Si = 2108 J/(Kg.degree C)

Lf = 334 *10^3 J/(kg)

heat absorbed by ice and water = m * (Lf + T1 * Si + Sw * T2)

heat absorbed by ice and water = 0.0159 * (334 *10^3 + 4186 * 28.2 + 2108 * 10.3)

heat absorbed by ice and water = 7532 J

the heat absorbed by ice and water is 7532 J