$37 \ g$ of ice at $0^{\circ} C$ is mixed with $74 \ g$ of water at $70^{\circ} C$. What is the resultant temperature (in $^{\circ} C$)? (Specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$)

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

$1\, kg$ of ice at $-10\,^{\circ}C$ is mixed with $4.4\, kg$ of water at $30\,^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Specific heat of ice = $2100\, J/kg\cdot K$,specific heat of water = $4200\, J/kg\cdot K$,latent heat of fusion of ice = $3.36 \times 10^5\, J/kg$)

The $SI$ unit of mechanical equivalent of heat is

$A$ 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, $C_{\text{water}} = 4190 \,J/kg^{\circ} C$, $C_{\text{ice}} = 2100 \,J/kg^{\circ} C$, $L_F = 3.34 \times 10^5 \,J/kg$]

$A$ liquid of mass $m$ and specific heat $c$ is at a temperature of $2T$. Another liquid of mass $m/2$ and specific heat $2c$ is at a temperature of $T$. What will be the final temperature of the mixture?