The specific heat of water is $4200 \, J \, kg^{-1} \, K^{-1}$ and the latent heat of ice is $3.4 \times 10^{5} \, J \, kg^{-1}$. $100 \, g$ of ice at $0^{\circ} C$ is placed in $200 \, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of the water reaches $0^{\circ} C$ is close to (in grams):

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 $= 540 \ cal/g$)

$A$ calorimeter has a heat capacity of $100 \ J/K$ and is at $30^{\circ}C$. $100 \ g$ of water at $40^{\circ}C$ (specific heat capacity $4200 \ J/kg \cdot K$) is poured into the calorimeter. What is the final temperature of the mixture in the calorimeter in $^{\circ}C$?

$A$ liquid of mass $m$ and specific heat $c$ is at a temperature of $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is at a temperature of $T$. What will be the final temperature of the mixture?

$A$ $10 \ W$ electric heater is used to heat a container filled with $0.5 \ kg$ of water. It is found that the temperature of the water and the container rises by $3 \ K$ in $15 \ min$. The container is then emptied,dried,and filled with $2 \ kg$ of oil. The same heater now raises the temperature of the container-oil system by $2 \ K$ in $20 \ min$. Assuming that there is no heat loss in the process and the specific heat of water is $4200 \ J \ kg^{-1} \ K^{-1}$,the specific heat of oil in the same unit is equal to: