Question

# To treat a burn on your hand, you decide to place an ice cube on the...

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 17.5 g, and its initial temperature is -11.1 °C. The water resulting from the melted ice reaches the temperature of your skin, 30.5 °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.

Given the mass of ice = 17.5 g

The heat absorbed by ice can be divided into 3 parts

Q1 = heat absorbed by ice at - 11.1 C to convert t ice at 0 C = mxs(ice)xdT

= 17.5 g x (2.06 J / gxC)x11.1 C = 400 J

Q2 = Heat absorbed by ice at 0 C to melt to water at 0 C = mxLf = 17.5 g x 333.55 J / g = 5837 J

Q3 = Heat absorbed by water to convert from water at 0 C to water at 30.5 C = mxs(water)xdT

= 17.5 g x (4.18 J / gxC)x30.5 C = 2231 J

Hence total heat absorbed,

Qt = Q1+Q2+Q3 = 400J + 5837 J + 2231 J = 8468 J (answer)

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