Question

Hot water of mass 10 kg at 90°C is cooled forr taking bath by mixing 20kg of water at 20°C. What will be the final temperature of the water neglecting the heat taken by bucket ?

Answer 1

Let the final temperature of the mixture be `t^@C`

Equating the heat loss and heat gained we get

`20xx4.2xx(t-20)=10xx4.2xx(90-t)`

`=>20t-400=900-10t`

`=>30t=1300`

`=>t=43.3^@C`