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

Q = heat change for conversion of ice at -10.7oC to ice at 0 oC + heat change for
conversion of ice at 0oC to water at 0oC + heat change for conversion of water at 0oC to water at 28.7 oC

Amount of heat absorbed ,

Q = mcdt + mL + mc'dt
= m(cdt + L + c'dt' )

Where
m = mass of ice =17.4 g
c' = Specific heat of water = 4.186 J/g degree C
c = Specific heat of ice= 2.09 J/g degree C
L= Heat of fusion of ice = 334.9 J/g
dt' = 28.5-0 =28.5 oC
dt = 0-(-10.7)=10.7 oC
Plug the values we get

Q = m(cdt + L + c'dt ' )= 8292.2J