Question

- 20g of cream (3°C) is added to 200g of coffee (100°C) in a thermally isolated container. If both the coffee and cream have the specific heat of water (4186 J/kgK), what is the final temperature?

Answer #1

The heat gained by the cream will be equal to the heat lost by the
coffee.

The heat gained by the cream,

H = m c
T = 20 x 10-3 kg x 4186 J/kg K x (T - 280 K)

The heat lost by the coffee,

H = m c
T =200 x 10-3 x 4186 J/kg K x (373 K - T )

Equating these two, we get

20 x 10-3 kg x 4186 J/kg K x (T - 280 K) = 200 x 10-3 x 4186 J/kg K
x (373 K - T )

20 x (T - 280 K) = 200 x (373 K - T )

T - 280 = 3130 - 10 T

11 T = 3410

T = 310 K

T = 370C

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