In an industrial process $10\, kg$ of water per hour is to be heated from $20^o C$ to $80^o C$ . To do this steam at $200^o C$ 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 $90^o C$. How many kg of steam is required per hour. (Specific heat of steam $= 0.5\, cal/g^o C$, Latent heat of vaporisation $= 540 cal/g)$

A piece of ice (heat capacity $=2100 \mathrm{~J} \mathrm{~kg}^{-1}{ }^{\circ} \mathrm{C}^{-1}$ and latent heat $=3.36 \times 10^5 \mathrm{~J} \mathrm{~kg}^{-1}$ ) of mass $\mathrm{m}$ grams is at $-5^{\circ} \mathrm{C}$ at atmospheric pressure. It is given $420 \mathrm{~J}$ of heat so that the ice starts melting. Finally when the ice-water mixture is in equilibrium, it is found that $1 \ \mathrm{gm}$ of ice has melted. Assuming there is no other heat exchange in the process, the value of $m$ is

Find the amount of heat supplied to decrease the volume of an ice water mixture by $1 \,\,cm^3$ without any change in temperature. $(\rho_ {ice} = 0.9 \rho_{water}, L_{ice} = 80 \,\,cal/gm).$ ……… $cal$

The specific heat of water $=4200\, J\, kg ^{-1}\, K ^{-1}$ and the latent heat of ice $=3.4 \times 10^{5}\, J\, kg ^{-1}.$ $100$ grams of ice at $0^{\circ} C$ is placed in $200\, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of water reaches $0^{\circ} C$ is close to (in $grams$) 

A copper block of mass $5.0\, kg$ is heated to a temperature of $500^{\circ} C$ and is placed on a large ice block. What is the maximum amount of ice (ઇન $kg$) that can melt? [Specific heat of copper: $0.39\, Jg ^{-1 \circ} C ^{-1}$ and latent heat of fusion of water : $335 \,J g ^{-1}$ ]