The heat required to raise the temperature of $50 \, g$ of copper by $10^\circ C$ is given to $10 \, g$ of water. What is the rise in temperature of the water in $^\circ C$? (Specific heat of copper $= 420 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$,Specific heat of water $= 4200 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$)

The temperatures of equal masses of three different liquids $A, B$ and $C$ are $15^{\circ} C, 24^{\circ} C$ and $30^{\circ} C$ respectively. The resultant temperature when liquids $A$ and $B$ are mixed is $20^{\circ} C$ and when liquids $B$ and $C$ are mixed is $26^{\circ} C$. Then the ratio of specific heat capacities of the liquids $A, B$ and $C$ is

The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$. The water is initially at a different temperature $T_W$. Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium,with no loss of heat to the surroundings. The final temperature of the mixture will be

Two liquids of equal mass have temperatures $32^{\circ}C$ and $24^{\circ}C$. When mixed,the final temperature becomes $28^{\circ}C$. What is the ratio of their specific heats?

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 $y/x$?