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 11.3 g, and its initial temperature is -14.4 °C. The water resulting from the melted ice reaches the temperature of your skin, 32.0 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand. Answer ______J

Constants may be found here:

Quantity	per gram	per mole
Enthalpy of fusion	333.6 J/g	6010. J/mol
Enthalpy of vaporization	2257 J/g	40660 J/mol
Specific heat of solid H2O (ice)	2.087 J/(g·°C) *	37.60 J/(mol·°C) *
Specific heat of liquid H2O (water)	4.184 J/(g·°C) *	75.37 J/(mol·°C) *
Specific heat of gaseous H2O (steam)	2.000 J/(g·°C) *	36.03 J/(mol·°C) *

Answer 1

The "hint" is very explicit and tells you exactly what to do. The ice goes through three changes to get from ice at -13.5C to water at 28.4C. You have to determine the amount of heat absorbed for each of these three changes and add them together to get the total energy absorbed by the "ice".

Changes:
1. Ice warms from 14.4C 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 32C. 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)]