Help With Chem HW

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 11.0g, and its initial temperature is  -11.6 C. The water resulting from the melted ice reaches the temperature of his skin, 32.0 C.  How much heat is absorbed by the ice cube and resulting water? Assume that all of the water remains in the hand.

Answer 1

To answer this question properly, one needs to know the specific heat of ice, the heat of fusion, and the specific heat of liquid water. The values I have for these are as follows:

specific heat of ice = 2.09 J/gº

heat of fusion of water = ∆Hfusion = 334 J/g

specific heat liquid water = 4.184 J/gº

To change temperature of 11.0 g of ice from -11.6º to 0º = q = mC∆T = (11.0g)(2.09 J/gº)(11.6º) = 266.68 J

To melt 11.0 g of ice at 0º = q = (m)(∆Hfusion) = (11.0 g)(334 J/g) = 3674 J

To raise temperature of 11.0 g liquid water from 0º to 32.0º = q = mC∆T = (11g)(4.184 J/gº)(32º) = 1472.77 J

Add up all the joules (3 sig. figs.) to find the amount of heat absorbed by the ice cube:

267 J + 3670 J + 1470 J = 5407 J = 5410 J (3 sig. figs.)