To treat a burn on your hand, you decide to place an ice cube on the burned skin.

Question

The mass of the ice cube is 12.2 g, and its initial temperature is -13.9 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.2 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand

Heat-transfer constants for H₂O at 1 atm

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

heat to change ice from -13.9º to 0º = mC∆T = (12.2g)(2.087 J/g/deg)(13.9 deg) = 354 J

heat to melt the ice @ 0º = m∆Hf = (12.2g)(333.6 J/g) = 4070 J

heat to change water from 0º to 31.2º = mC∆T = (12.2g)(4.184 J/g/deg)(31.2 deg) = 1593 J

Add up all the heats to obtain 6017 J of heat absorbed by the ice cube