In an industrial process,$10 \, kg$ of water per hour is to be heated from $20^\circ C$ to $80^\circ C$. To do this,steam at $150^\circ C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^\circ C$. How many $kg$ of steam is required per hour? (Specific heat of steam $= 1 \, cal/g^\circ C$,Latent heat of vaporisation $= 540 \, cal/g$)

$A$ small quantity of water of mass $m$ at temperature $\theta^{\circ} C$ is mixed with a large mass $M$ of ice which is at its melting point. If $s$ is the specific heat capacity of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is:

$A$ $10 \ W$ 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:

$20 \, g$ of boiling water is poured into an ice-cold brass vessel (specific heat $0.1 \, cal/g-^{\circ}C$) of mass $100 \, g$. The resulting temperature is ........ $^{\circ}C$.

$M$ 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:

Steam at $100\,^{\circ}C$ is passed into $20\,g$ of water at $10\,^{\circ}C$. When water acquires a temperature of $80\,^{\circ}C$,the mass of water present will be ........ $g$. [Take specific heat of water $= 1\,cal\,g^{-1}\,^{\circ}C^{-1}$ and latent heat of steam $= 540\,cal\,g^{-1}$]