One kilogram of ice at $0\,^{\circ}C$ is mixed with one kilogram of water at $80\,^{\circ}C$. The temperature of the mixture is ........ $^{\circ}C$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,kJ\,kg^{-1}$).

Steam at $100^{\circ}C$ is passed into $1.1 \, kg$ of water contained in a calorimeter of water equivalent $0.02 \, kg$ at $15^{\circ}C$ until the temperature of the calorimeter and its contents rises to $80^{\circ}C$. The mass of the steam condensed in $kg$ is:

Two liquids $A$ and $B$ are at temperatures of $75^{\circ} C$ and $15^{\circ} C$. Their masses are in the ratio of $2: 3$ and their specific heats in the ratio $3: 4$. If these two liquids are mixed,then the resulting temperature will be $...^{\circ} C$.

The time required to raise the temperature of $3 \text{ litre}$ of water from $0^{\circ} C$ to $80^{\circ} C$ by a heater operated under $200 \text{ V}$ having resistance of $50 \Omega$ is
[specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$] [density of water $= 1000 \text{ kg/m}^3$] (in $\text{ min}$)

We have half a bucket $(6 \text{ litre})$ of water at $20^{\circ}C$. If we want water at $40^{\circ}C$,how much steam at $100^{\circ}C$ should be added to it?

$A$ 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: