Question

3. Hot water of mass 300 g and at a temperature of 90 °C is added to 200 g of cold water at 10 °C. What is the final temperature of the mixture?

Answer 1

58°C

Explanation:

Heat absorbed by hot water + heat absorbed by cold water = 0

m₁C₁ΔT₁ + m₂C₂ΔT₂ = 0

(300 g) (4.184 J/g/K) (T − 90°C) + (200 g) (4.184 J/g/K) (T − 10°C) = 0

(300 g) (T − 90°C) + (200 g) (T − 10°C) = 0

(300 g) (T − 90°C) = -(200 g) (T − 10°C)

3 (T − 90°C) = -2 (T − 10°C)

3T − 270°C = -2T + 20°C

5T = 290°C

T = 58°C