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 16.4 g, and its initial temperature is −13.6 ∘C. The water resulting from the melted ice reaches the temperature of his skin, 31.7 ∘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

Answer 1

Answer:

8115.88 J

Explanation:

$m$ = mass of ice = 16.4 g

$c_{i}$ = specific heat of ice = 2.087 J/(g °C)

$T_{ii}$ = initial temperature of ice = - 13.5 °C

$T_{s}$ = Skin temperature = 31.7 °C

$L$ = latent heat of fusion of ice to water = 334 J/g

$c_{w}$ = specific heat of water = 4.186 J/(g °C)

heat absorbed is given as

$Q = m c_{i} (0 - T_{ii}) + m L + m c_{w} (T_{s}- 0)$

$Q = (16.4) (2.087) (0 - (- 13.5)) + (16.4) (334) + (16.4) (4.186) (31.7 - 0)$

$Q$ = 8115.88 J