$A$ piece of metal of mass $5 \ kg$ and temperature $-50 \ ^\circ C$ is dropped in $20 \ kg$ water at $52 \ ^\circ C$ temperature. In thermal equilibrium,the temperature of water decreases by $2 \ ^\circ C$. Find the specific heat of the metal in $Cal/g \ ^\circ C$.

$200 \, g$ of a solid ball at $20 \, ^\circ C$ is dropped in an equal amount of water at $80 \, ^\circ C$. The resulting temperature is $60 \, ^\circ C$. This means that the specific heat of the solid is

When $50 \ g$ of water at $10^{\circ} C$ is mixed with $50 \ g$ of water at $100^{\circ} C$,the resultant temperature is: (in $^{\circ} C$)

$0.1 \,m^{3}$ of water at $80^{\circ} C$ is mixed with $0.3 \,m^{3}$ of water at $60^{\circ} C$. The final temperature of the mixture is (in $^{\circ} C$)

$A$ beaker contains $200\,g$ of water. The heat capacity of the beaker is equal to that of $20\,g$ of water. The initial temperature of water in the beaker is $20\,^{\circ}C$. If $440\,g$ of hot water at $92\,^{\circ}C$ is poured in it,the final temperature (neglecting radiation loss) will be nearest to ........ $^{\circ}C$.

$540 \ g$ of ice at $0 \ ^\circ C$ is mixed with $540 \ g$ of water at $80 \ ^\circ C$. The final temperature of the mixture is