Question

When 1 kg of gold bar of specific heat capacity 129 J/kg°C at 100°C is dropped into 1 kg of water at 25°C, the resulting temperature is:

Answer 1

@ParthKohli

Answer 2

let final temperature is T
heat loss by gold bar= heat gain by water
\[m _{g} C _{g} (100-T)=m _{w} C _{w}(T-25)\]