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 $= 540 \, cal/g$)

The temperature of water of mass $100 \ g$ is raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam to it. The mass of the steam added is (Latent heat of steam $= 540 \ cal \ g^{-1}$ and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

The water equivalent of a calorimeter is $10 \ g$ and it contains $50 \ g$ of water at $15^{\circ} C$. Some amount of ice,initially at $-10^{\circ} C$,is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (given specific heat of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,specific heat of water $= 1.0 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of melting of ice $= 80 \ cal \ g^{-1}$) (in $g$)?

$M$ 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:

$10 \, g$ of ice at $0 \, ^oC$ is mixed with $m \, g$ of water at $50 \, ^oC$. What is the minimum value of $m$ (in $g$) so that the ice melts completely? (Given: $L_f = 80 \, cal/g$ and $S_W = 1 \, cal/g \cdot ^oC$)

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