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 11.4 g, and its initial temperature is − 11.1 ∘ C. The water resulting from the melted ice reaches the temperature of his skin, 28.0 ∘ 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

Specific heat of ice = 2.03 J/g^oC

Specific heat of water = 4.18 J/g^oC

Heat of fusion of ice = 334 J/g

Heat absorbed by ice in reaching 0 ^oC from -11.1 ^oC = mass of ice x specific heat of ice x temperature difference

                                                                                    = 11.4 g x 2.03 J/g^oC x (0 - (-11.1))^oC

                                                                                    = 257 J

Heat absorbed on ice melting = mass of ice x heat of fusion of ice

                                                = 11.4 g x 334 J/g

                                                = 3808 J

Heat absorbed by water in reaching the temperature from 0 ^oC to 28.0 ^oC = mass of water x specific heat of water x temperature difference

                                                                                    = 11.4 g x 4.18 J/g^oC x (28 - 0)^oC

                                                                                    = 1334 J

Hence, the total heat absorbed by ice and water = (257 + 3808 + 1334) J

                                                                             = 5339 J