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

Steam at $100^{\circ} C$ is passed into $1 \ kg$ of water contained in a calorimeter of water equivalent $0.2 \ kg$ at $9^{\circ} C$ until the temperature of the calorimeter and water in it increases to $90^{\circ} C$. The mass of steam condensed in $kg$ is nearly (specific heat of water $= 1 \ cal/g^{\circ} C$,latent heat of vaporisation $= 540 \ cal/g$)

$37 \ g$ 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 $^{\circ} C$)

$19 \ g$ of water at $30^{\circ} C$ and $5 \ g$ of ice at $-20^{\circ} C$ are mixed together in a calorimeter. What is the final temperature of the mixture (in $^{\circ} C$)? Given specific heat of ice $= 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$.

If an electric heater is rated at $1000\, W$, then the time required to heat one litre of water from $20\, ^oC$ to $60\, ^oC$ is

$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$,specific heat of water $= 4200 \ J/kg.^{\circ}C$,latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)