$A$ calorimeter contains $0.2 \ kg$ of water at $30^{\circ}C$. When $0.1 \ kg$ of water at $60^{\circ}C$ is added to it,the final temperature of the mixture becomes $35^{\circ}C$. What is the heat capacity of the calorimeter in $J/K$?

$A$ piece of copper of mass $250 \ g$ at $500^{\circ} C$ is put inside a calorimeter of water equivalent $50 \ g$ containing $200 \ g$ of water at $20^{\circ} C$. At thermal equilibrium,the temperature of the mixture is $60^{\circ} C$. The specific heat of copper (in $J / kg \cdot ^{\circ} C$) is approximately: [Specific heat of water $= 4200 \ J / kg \cdot ^{\circ} C$]

Two identical bodies are made of a material for which the heat capacity increases with temperature. One of these is held at a temperature of $100^{\circ} C$,while the other one is kept at $0^{\circ} C$. If the two are brought into contact,then assuming no heat loss to the environment,the final temperature that they will reach is

$50\, g$ of ice at $0\,^{\circ}C$ is dropped into a calorimeter containing $100\, g$ of water at $30\,^{\circ}C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is ........ $g$.

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

$M$ grams of steam at $100^{\circ} C$ is mixed with $200 \; g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 \; cal/g$ and heat of fusion of ice is $80 \; cal/g$],the value of $M$ is: