Question

You pour 200 g hot coffee at 78.7°C and some cold cream at 7.50°C to a 115-g cup that is initially at a temperature of 22.0°C. The cup, coffee, and cream reach an equilibrium temperature of 62.0°C. The material of the cup has a specific heat of 0.2604 kcal/(kg · °C) and the specific heat of both the coffee and cream is 1.00 kcal/(kg · C). If no heat is lost to the surroundings or gained from the surroundings, how much cream did you add?

Answer 1

The overall change in energy must be 0.

Change in energy of the coffee

= (0.200 kg) * (62.0 - 78.7 C°) * (1.00 kcal/(kg C°) )

= -3.34 kcal

Change in energy of the cream

= (x kg) * (62.0 - 7.50 C°) * (1.00 kcal/(kg C°) )

= 54.5x kcal

Change in energy of the cup

= (0.115 kg) * (62.0 - 22.0 C°) * (0.2604 kcal/(kg C°))

= 1.19784 kcal

If we add them up, the sum should be 0 (since no heat was lost to the surroundings):

-3.34 + 54.5x + 1.19784 = 0

x = 0.0393

x = 0.039 ( appprox)

Therefore, the amount of cream was

x = 0.039kg = 39 g

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