Question

1 kg of water (C = 4.18 J/g K) at 50^o is mixed with 0.2 kg of ethanol (C = 2.46J/g K) at 0^oC. What is the final temperature?

An ice block whose mass is 2 kg is at 0^oC. It is put into a container with 10 kg of water at 50^oC. What is the final temperature?

Answer 1

part A

(4.18 J/g) x (1000 g) = 4180 J gained by the ice and lost by the water

Calculate the temperature of the mixer after water is mixed
(4180 J) / (2.46 J/(g·K) ) / (200 g) = 8.49 °C change
50°C - 8.49°C = 41.50°C

So now you have 1 kg liquid water at 50°C and 0.2 kg ethanol at 0°C. When they are mixed:
(1 kg) x (50°C) + (0.2 kg) x (41.50°C) / (1 kg + 0.2 kg) = 48.58 °C

part B

(333.6 J/g) x (2000 g) = 667200 J gained by the ice and lost by the water

Calculate the temperature of the warm water after the heat that melted the ice is removed:
(667200 J) / (4.184 J/(g·°C) ) / (10000 g) = 15.95 °C change
50°C - 15.95°C = 34.05°C

So now you have 2 kg liquid water at 0°C and 10 kg liquid water at 34.05°C. When they are mixed:
(2 kg) x (0°C) + (10 kg) x (34.05°C) / (2 kg + 10 kg) = 28.4 °C