When $50 \ g$ of water at $10^{\circ} C$ is mixed with $50 \ g$ of water at $100^{\circ} C$,the resultant temperature is: (in $^{\circ} C$)

The temperature of a copper piece of mass $50 \ g$ is raised by $10 \ ^\circ C$. If the same amount of heat is given to $10 \ g$ of water,the rise in its temperature is = ...... $^\circ C$ (Specific heat of copper $= 420 \ J/kg \cdot ^\circ C$,Specific heat of water $= 4200 \ J/kg \cdot ^\circ C$).

When $x \text{ grams}$ of steam at $100^{\circ}C$ is mixed with $y \text{ grams}$ of ice at $0^{\circ}C$,we obtain $(x + y) \text{ grams}$ of water at $100^{\circ}C$. What is the ratio $y/x$?

$10 \, kg$ 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 $= 540 \, cal/g$)

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 $\text{ min}$)

$1\, g$ of steam at $100^{\circ}C$ melts ........ $g$ of ice at $0^{\circ}C?$ (Latent heat of ice $= 80\, cal/g$ and latent heat of steam $= 540\, cal/g$)