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An ice cube with a mass of 46.4 g at 0.0 ∘C is added to a...

An ice cube with a mass of 46.4 g at 0.0 ∘C is added to a glass containing 4.20×102 g of water at 45.0 ∘C . Determine the final temperature of the system at equilibrium. The specific heat capacity of water, ?s , is 4.184 J/g⋅∘C , and the standard enthalpy of fusion, Δ?∘fus , of water is 6.01×103 J/mol . Assume that no energy is transferred to or from the surroundings.

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Answer #1

mass of ice cube = 46.4 g

moles of ice = 46.4 / 18.015 = 2.575

temperature = 0.0 oC

mass of water = 420 g

temperature = 45 oC

heat absorbed by ice cube = heat release by water

Q ice = 2.575 x 6.01 x 10^3 + 46.4 x 4.184 x (Tf - 0)

         = 15479.54 + 194.14 (Tf - 0)

heat released = 420 x 4.184 x (45 - Tf)

                      = 1757.28 x (45 - Tf)

1757.28 x (45 - Tf) = 15479.54 + 194.14 (Tf - 0)

79077.6 - 1757.28 Tf = 15479.54 + 194.14 Tf

Tf = 32.59

final temperature = 32.6 oC

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