Question

A 40 g block of ice is cooled to -69°C. and is then added to 630...

A 40 g block of ice is cooled to -69°C. and is then added to 630 g of water in an 80 g copper calorimeter at a temperature of 27°C. Determine the final temperature of the system consisting of the ice, water, and calorimeter. Remember that the ice must first warm to 0°C, melt, and then continue warming as water. The specific heat of ice is 0.500 cal/g ·°C = 2090 J/kg°C.

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Answer #1

Let m1 be the mass of ice, m2 be the melt, m3 be the cold water, m4 be the warm water & m5 be mass of copper, respectively.

Q1 + Q2 + Q3 + Q4 + Q5 = 0

m1 Cice [0 - (-69 0C)] + m2H + m3 Cw (Tf - Ti,C) + m4 Cw (Tf - Ti,H) + m5 Ccu (Tf - Ti,H) = 0

where, Ti,C = 0

(0.04 kg) (4186 J/kg 0C) Tf + (0.63 kg) (4186 J/kg 0C) (Tf - Ti,H) + (0.08 kg) (387 J/kg 0C) (Tf - Ti,H) = - (0.04 kg) (2090 J/kg 0C) (69 0C) - (0.04 kg) (333000 J/kg)

(167.4 J/0C) Tf + (Tf - Ti,H) [(2637.1 J/0C) + (30.9 J/0C)] = - (5768.4 J) - (13320 J)

(167.4 J/0C) Tf + (Tf - Ti,H) [(2637.1 J/0C) + (30.9 J/0C)] = - (19088.4 J)

(167.4 J/0C) Tf + (2668 J/0C) [Tf - (27 0C)] = - (19088.4 J)

(167.4 J/0C) Tf + (2668 J/0C) Tf = - (19088.4 J) + (72036 J)

Tf = (52947.6 J) / (2835.4 J/0C)

Tf = 18.6 0C

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