$A$ 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 $Cal/g \ ^\circ C$.

The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$. The water is initially at a different temperature $T_W$. Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium,with no loss of heat to the surroundings. The final temperature of the mixture will be :-

$A$ piece of metal weighing $100 \,g$ is heated to $80^{\circ} C$ and dropped into $1 \,kg$ of cold water in an insulated container at $15^{\circ} C$. If the final temperature of the water in the container is $15.69^{\circ} C$,the specific heat of the metal in $J / g^{\circ} C$ is:

$200 \, g$ 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

Steam at $100\,^{\circ}C$ is passed into $20\,g$ of water at $10\,^{\circ}C$. When water acquires a temperature of $80\,^{\circ}C$,the mass of water present will be ........ $g$. [Take specific heat of water $= 1\,cal\,g^{-1}\,^{\circ}C^{-1}$ and latent heat of steam $= 540\,cal\,g^{-1}$]

The thermal capacity of a body is $80 \, cal/^{\circ}C$. What is its water equivalent?