Time taken by a $836\; W$ heater to heat $1\; litre$ of water from $10^{\circ} C$ to $40^{\circ} C$ is (in $; s$)

The time required to raise the temperature of $3 \text{ litre}$ of water from $0^{\circ} C$ to $80^{\circ} C$ by a heater operated under $200 \text{ V}$ having resistance of $50 \Omega$ is
[specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$] [density of water $= 1000 \text{ kg/m}^3$] (in $\text{ min}$)

How many grams of ice at $0 \, ^\circ \text{C}$ will be melted by $1 \, \text{g}$ of steam at $100 \, ^\circ \text{C}$? (Latent heat of fusion of ice $L = 80 \, \text{cal/g}$ and latent heat of vaporization of water $L' = 540 \, \text{cal/g}$)

$19 \ g$ 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 $= 80 \ cal \ g^{-1}$.

$A$ calorimeter of water equivalent $20 \, g$ contains $180 \, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it until the temperature of the mixture is $31^{\circ} C$. The value of '$m$' is close to (Latent heat of water $= 540 \, \text{cal} \, g^{-1}$,specific heat of water $= 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1}$)

$5 \ kg$ of water at $20^{\circ} C$ is added to $10 \ kg$ of water at $60^{\circ} C$. Neglecting the heat capacity of the vessel and other losses,the resultant temperature will be nearly: (in $^{\circ} C$)