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 13.2 g, and its initial temperature is -12.6 °C. The water resulting from the melted ice reaches the temperature of your skin, 28.2 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand. Constants may be found here.

Answer 1

Ti = -12.6

Tf = 28.2

here

Cs = 2.09 J/goC

Heat required to convert solid from -12.6 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 13.2 g * 2.09 J/goC *(0--12.6) oC

= 347.6 J

Lf = 333.0 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 13.2g *333.0 J/g

= 4395.6 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 28.2 oC

Q3 = m*Cl*(Tf-Ti)

= 13.2 g * 4.184 J/goC *(28.2-0) oC

= 1557.5 J

Total heat required = Q1 + Q2 + Q3

= 347.6 J + 4395.6 J + 1557.5 J

= 6301 J