Question

1. 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

Let T be the final temperature.
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