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 15.215.2 g, and its initial temperature is −14.6−14.6 °C. The water resulting from the melted ice reaches the temperature of your skin, 31.931.9 °C. How much heat is absorbed by the ice cube and resulting water? Assume that all the water remains in your hand.

Answer 1

Ti = -14.6 oC

Tf = 31.9 oC

here

Cs = 2.087 J/goC

Heat required to convert solid from -14.6 oC to 0.0 oC

Q1 = m*Cs*(Tf-Ti)

= 15.2 g * 2.087 J/goC *(0--14.6) oC

= 463.147 J

Lf = 333.6 J/g

Heat required to convert solid to liquid at 0.0 oC

Q2 = m*Lf

= 15.2g *333.6 J/g

= 5070.72 J

Cl = 4.184 J/goC

Heat required to convert liquid from 0.0 oC to 31.9 oC

Q3 = m*Cl*(Tf-Ti)

= 15.2 g * 4.184 J/goC *(31.9-0) oC

= 2028.7379 J

Total heat required = Q1 + Q2 + Q3

= 463.147 J + 5070.72 J + 2028.7379 J

= 7563 J

= 7.563 KJ

Answer: 7.563 KJ