Question

200 g of hot water at 80°C is added to 300 g of cold water at 10 °C. Calculate the final temperature of the mixture of water. Consider the heat taken by the container to be negligible. [specific heat capacity of water is 4200 J kg^-1 °C^-1]

Answer 1

Given Hot water m = 200 g, temperature = 80°C

Cold water m = 300 g, temperature = 10°C

Let final temperature of mixture = θ

By the principle of calorimeter

Heat given = Heat taken

200 × c × (80 - θ) = 300 × c × (θ - 10)

200 × 80 - 200θ = 300 θ  - 300 × 10

16000 - 200 θ = 300 θ - 3000

19000 = 500 θ

θ = 38°C