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 $\,g$)

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

$37 \ g$ of ice at $0^{\circ} C$ is mixed with $74 \ g$ of water at $70^{\circ} C$. What is the resultant temperature (in $^{\circ} C$)? (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$)

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

$A$ massless spring $(k = 800\, N/m)$,attached with a mass $(500\, g)$ is completely immersed in $1\, kg$ of water. The spring is stretched by $2\, cm$ and released so that it starts vibrating. What would be the order of magnitude of the change in the temperature of water when the vibrations stop completely? (Assume that the water container and spring receive negligible heat and specific heat of mass $= 400\, J/kg\, K$,specific heat of water $= 4184\, J/kg\, K$)

$37 \ g$ of ice at $0^{\circ} C$ temperature is mixed with $74 \ g$ of water at $70^{\circ} C$ temperature. The resultant temperature is (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$) (in $^{\circ} C$)