We have half a bucket $(6\,L)$ of water at $20\,^oC$. If we want water at $40\,^oC$,how much steam at $100\,^oC$ should be added to it?

The temperature of $100 \,g$ of water is to be raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam at $100^{\circ} C$ to it. The mass of the steam required in this process is (latent heat of steam is $540 \,cal \,g^{-1}$). (in $\,g$)

Steam at $100^{\circ} C$ is passed into $114 \ g$ of water at $30^{\circ} C$. The mass of water present in the mixture when the temperature of the mixture becomes $70^{\circ} C$ is (Latent heat of steam $= 540 \ cal \ g^{-1}$; specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

What is calorimetry? What is a calorimeter? Explain its principle and construction.

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is: