Question

The temperature of 600 g of cold water rises by 15^°C when 300 g of hot water at 50^°C is added to it. What was the initial temperature of the cold water?

Answer 1

Let final temp. of mixture = T^°C

∴ Rise in temp. of cold water = T - t = 15^°C

Let C be the Specific heat capacity of water

Heat lost by hot water = Heat gained by cold water

Mass × C × fall in temperature = Mass × C × Rise in temperature

⇒ 300 × C × (50 - T) = 600 × C × (T - t)

⇒ 50 - T = 2 × [15]

⇒ T = 50 - 30 = 20

But rise in temperature = T - initial temperature of cold water

⇒ 15 = 20 - t

∴ t = 20 - 15 = 5^°C