Question

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

Answer 1

Heat gained by cold water = Heat lost by hot water

$\Rightarrow mS\Delta T=m^{\prime }S\Delta T^{\prime }\Rightarrow m\Delta T=m^{\prime }\Delta T^{\prime }\Rightarrow 600\times 15=300\times \Delta T^{\prime }\Rightarrow \Delta T^{\prime }={30}^{o}C$

Change in temperature of hot water is 30°C. Let the final temperature of mixture be T.

So, 30°C = 50°C - T ………[∵ Initial Temperature of water was 50°C]

$\Rightarrow$ T = 20°C

Given that, rise in temperature of cold water = 15°C

$\Rightarrow$ 20°C - T’ = 15°C

$\Rightarrow$ T’ = 5°C

∴ Initial Temperature of cold water is 5°C.