Two tanks $A$ and $B$ contain water at $30\,^oC$ and $80\,^oC$ respectively. Calculate the amount of water that must be taken from each tank to prepare $40\,kg$ water at $50\,^oC$

When $x\, grams$ of steam at $100\,^oC$ is mixed  with $y\,grams$ of ice at $0\,^oC$ , We obtain $(x + y)\,grams$ of water at $100\,^oC$  . What is the ratio $y/x$ ?

When $0.15\; kg$ of $1 ce$ at $0^{\circ} C$ is mixed with $0.30 \;kg$ of water at $50^{\circ} C$ in a container, the resulting temperature is $6.7^{\circ} C$. Calculate the heat of fuston of ice. $(s_{\text {water }}=4186 J kg ^{-1} K ^{-1}$ ).

Ice at $-20\,^oC$ is added to $50\,g$ of water at $40\,^oC.$ When the temperature of the mixture reaches $0\,^oC,$ it is found that $20\,g$ of ice is still unmelted. The amount of ice added to the water was close to ........$g$ (Specific heat of water $= 4.2\,J/g/^oC)$ Heat of fusion of water at $0^oC = 334\,J/g$ )

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

$250\,gm$ of water and an equal volume of alcohol of mass $200\,gm$ are placed successively in the same calorimeter and cools from $60^{\circ}\,C$ to $55^{\circ}\,C$ in $130\,sec$ and $67 sec$ respectively. If the water equivalent of the calorimeter is $10\,gm$. , then the specific heat of alcohol in cal/gm $cal / gm ^{\circ}\,C$ is