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 17.5 g, and its initial temperature is -11.1 °C. The water resulting from the melted ice reaches the temperature of your skin, 30.5 °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

Given the mass of ice = 17.5 g

The heat absorbed by ice can be divided into 3 parts

Q1 = heat absorbed by ice at - 11.1 C to convert t ice at 0 C = mxs(ice)xdT

= 17.5 g x (2.06 J / gxC)x11.1 C = 400 J

Q2 = Heat absorbed by ice at 0 C to melt to water at 0 C = mxLf = 17.5 g x 333.55 J / g = 5837 J

Q3 = Heat absorbed by water to convert from water at 0 C to water at 30.5 C = mxs(water)xdT

= 17.5 g x (4.18 J / gxC)x30.5 C = 2231 J

Hence total heat absorbed,

Qt = Q1+Q2+Q3 = 400J + 5837 J + 2231 J = 8468 J (answer)