$A$ metal ball of mass $100 \ g$ at $20^{\circ} C$ is dropped in $200 \ ml$ of water at $80^{\circ} C$. If the resultant temperature is $70^{\circ} C$,then the ratio of specific heat of the metal to that of water is

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

The temperature of $100 \, g$ of water is to be raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam at $100^{\circ} C$ to it. The mass of the steam required for this purpose is ........... $g$.

$15 \text{ g}$ of ice at $0^{\circ}\text{C}$ is added to a vessel containing water at $40^{\circ}\text{C}$. The mass of water and water equivalent of the vessel is $60 \text{ g}$. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be $[L_{\text{ice}} = 80 \text{ cal/g}, S_{\text{water}} = 1 \text{ cal/g}]$ (in $^{\circ}\text{C}$)

Steam at $100^{\circ} C$ is added to $150 \,g$ of water to increase its temperature from $20^{\circ} C$ to $40^{\circ} C$. The total mass of the water at $40^{\circ} C$ is (specific heat capacity of water $= 1 \,cal \,g^{-1} {}^{\circ} C^{-1}$ and latent heat of steam $= 540 \,cal \,g^{-1}$) (in $\,g$)

$A$ block of ice at $-20\,^oC$ having a mass of $2\,kg$ is added to $3\,kg$ of water at $15\,^oC$. Neglecting heat losses and the heat capacity of the container,what is the final state of the system?