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 ........... $g$.

When $100\,g$ of a liquid $A$ at $100\,^oC$ is added to $50\,g$ of a liquid $B$ at temperature $75\,^oC$,the temperature of the mixture becomes $90\,^oC$. The temperature of the mixture,if $100\,g$ of liquid $A$ at $100\,^oC$ is added to $50\,g$ of liquid $B$ at $50\,^oC$,will be ........$^oC$

$540 \ g$ of ice at $0 \ ^\circ C$ is mixed with $540 \ g$ of water at $80 \ ^\circ C$. The final temperature of the mixture is

$A$ calorimeter contains $0.2\, kg$ of water at $30\,^{\circ}C$. $0.1\, kg$ of water at $60\,^{\circ}C$ is added to it. The mixture is well stirred and the resulting temperature is found to be $35\,^{\circ}C$. The thermal capacity of the calorimeter is .......... $J/K$.

$10 \, kg$ of water is to be heated from $20^{\circ}C$ to $80^{\circ}C$ in $1 \, hour$. Steam at $150^{\circ}C$ is passed through a copper coil immersed in the water. The steam condenses and returns to the boiler at $90^{\circ}C$. How much steam is required per hour? (Specific heat of water $= 1 \, cal/g^{\circ}C$,Latent heat of vaporization $= 540 \, cal/g$)

$10 \ kg$ 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 $= 3.36 \times 10^{5} \ J/kg$)