General Chemistry

Question

To treat a burn on his hand, a person decides to place an ice cube on the burned skin. The mass of the ice cube is  15.1g and its initial temperature is  -13.6 C. The water resulting from the melted ice reaches the temperature of his skin,  29.5 C. How much heat is absorbed by the ice cube and resulting water? Assume that all of the water remains in the hand. Constants for water can be found in this.

q= j

Answer 1

Assuming all of the ice (15.1 g) has melted, then...

heat to raise temperature ice from -13.6 to 0 degrees = q = mC∆T = (15.1 g)(2.09 J/gº)(13.6º) = 429 J

heat to melt the ice at 0ºC = q = m∆Hf = (15.1 g)(334 J/g) = 5043 J

heat to change temp of 15.5 g H₂O(l) from 0ºC to 29.5ºC = q = mC∆T = (15.1 g)(4.184 J/gº)(29.5º) = 1864 J

Total heat absorbed by ice and liquid water = 429 J + 5043 J + 1864 J = 7335 J (7.34 kJ)