You pour 160 g hot coffee at 78.7°C and some cold cream at 7.50°C to a 115-g cup

Question

You pour 160 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 57.0°C. The material of the cup has a specific heat of 1091 J/(kg · K) and the specific heat of both the coffee and cream is 4190 J/(kg · K). Assume that no heat is lost to the surroundings or gained from the surroundings. (b) What is the absolute value of the heat transferred to the cream? J (c) How much cream was added? g

Explanation / Answer

The overall change in energy must be 0.

Change in energy of the coffee
= (0.160 kg) * (57.0 - 78.7 C°) * (4190 j/(kg K) )
= -14547.68 J

Change in energy of the cream
= (x kg) * (57.0 - 7.50 C°) * (4190 j/(kg.K) )
= 207405x J

Change in energy of the cup
= (0.115 kg) * (57.0 - 22.0 C°) * (1091 j/(kg K))
= 4391.275 J.

If we add them up, the sum should be 0 (since no heat was lost to the surroundings):
--14547.68 J + 207405x J + 4391.275 J = 0
x = 0.04896.

Correcting for significant figures in the calculation of x:
x = 0.049

That is, the amount of cream was
0.049 kg = 49 g.

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