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 $= 4200 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$)

Three containers $C_{1}, C_{2}$ and $C_{3}$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in litres) are taken from each container and mixed (assume no loss of heat during the process).

$C_{1}$	$C_{2}$	$C_{3}$	$T$
$1 \ l$	$2 \ l$	$-$	$60^{\circ} C$
$-$	$1 \ l$	$2 \ l$	$30^{\circ} C$
$2 \ l$	$-$	$1 \ l$	$60^{\circ} C$
$1 \ l$	$1 \ l$	$1 \ l$	$\theta$

The value of $\theta$ (in $^{\circ} C$ to the nearest integer) is

$2 \ g$ of steam at $100 \ ^\circ C$ is passed into $40 \ g$ of water at $25 \ ^\circ C$. The final temperature of the mixture becomes $54.3 \ ^\circ C$. Find the latent heat of vaporization of steam in $cal/g$.

In an experiment,a sphere of aluminium of mass $0.20\, kg$ is heated up to $150\, ^\circ C$. Immediately,it is put into water of volume $150\, cc$ at $27\, ^\circ C$ kept in a calorimeter of water equivalent $0.025\, kg$. The final temperature of the system is $40\, ^\circ C$. The specific heat of aluminium is ............ $J/kg\cdot ^\circ C$ (take $4.2\, J = 1\, cal$).

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 $g$)

$A$ glass beaker contains $200 \,g$ of carbonated water initially at $20^{\circ} C$. How much ice should be added to obtain the final temperature of $0^{\circ} C$ with all ice melted, if the initial temperature of ice is $-10^{\circ} C$ (in $\,g$)? Neglect the heat capacity of the glass.
[Take, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]