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 16.5 g, and its initial temperature is -11.9 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.9 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.

Answer 1

1. Ice warms from -11.9C to 0C, its melting point. We use q1 = mcDT to compute q1 since it underwent a temperature change only. Specific heat of ice is 2.11 J/gC.
2. Ice melts at 0C. This is a phase change. q2 = mHf ... where Hf is the heat of fusion (fusion = melting). For water this is 334 J/g. This converts the ice to liquid water, but with the same mass as the ice originally had.
3. Water (from ice) warms from 0C to 31.9 oC. q3 = mcDT. Again we have a temperature change. The specific heat of water is 4.18 J/gC. The mass is the same as the original ice.

q(total) = q1 + q2 + q3
or
q(total) = m(ice)[c(ice)DT(ice) + Hf + c(water)DT(water)]

q(total) = 16.5*2.11*11.9 + 16.5*333.6 + 16.5*4.186*31.9

q(total) = 8121.99 J/gC