How many grams of a liquid with a specific heat of $0.2 \, cal/g^{\circ}C$ at a temperature of $40^{\circ}C$ must be mixed with $100 \, g$ of a liquid with a specific heat of $0.5 \, cal/g^{\circ}C$ at a temperature of $20^{\circ}C$,so that the final temperature of the mixture becomes $32^{\circ}C$?

$A$ beaker contains $200\,g$ of water. The heat capacity of the beaker is equal to that of $20\,g$ of water. The initial temperature of water in the beaker is $20\,^{\circ}C$. If $440\,g$ of hot water at $92\,^{\circ}C$ is poured in it,the final temperature (neglecting radiation loss) will be nearest to ........ $^{\circ}C$.

Steam at $100\,^{\circ}C$ is passed into $22\,g$ of water at $20\,^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90\,^{\circ}C$ (Latent heat of steam is $540\,cal/g$) is ......... $g$.

Steam at $100^{\circ} C$ is passed into $1 \ kg$ of water contained in a calorimeter of water equivalent $0.2 \ kg$ at $9^{\circ} C$ until the temperature of the calorimeter and water in it increases to $90^{\circ} C$. The mass of steam condensed in $kg$ is nearly (specific heat of water $= 1 \ cal/g^{\circ} C$,latent heat of vaporisation $= 540 \ cal/g$)

$19 \ g$ of water at $30^{\circ} C$ and $5 \ g$ of ice at $-20^{\circ} C$ are mixed together in a calorimeter. What is the final temperature of the mixture (in $^{\circ} C$)? Given specific heat of ice $= 0.5 \ cal \ g^{-1} (^{\circ} C)^{-1}$ and latent heat of fusion of ice $= 80 \ cal \ g^{-1}$.

$A$ metal ball of mass $100 \ g$ at $20^{\circ} C$ is dropped in $200 \ ml$ of water at $80^{\circ} C$. If the resultant temperature is $70^{\circ} C$,then the ratio of specific heat of the metal to that of water is