$A$ $210 \,W$ heater is used to heat $100 \,g$ of water. The time required to raise the temperature of this water from $25^{\circ} C$ to $100^{\circ} C$ is (specific heat capacity of water $= 4200 \,J / kg \cdot ^{\circ} C$). (in $\,s$)

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

$2\,kg$ of metal at $100\,^{\circ}C$ is cooled by $1\,kg$ of water at $0\,^{\circ}C$. If the specific heat capacity of the metal is $\frac{1}{2}$ of the specific heat capacity of water,the final temperature of the mixture would be:

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 $\text{ min}$)

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

$1 \text{ kg}$ of steam at $150^{\circ} \text{C}$ is passed from a steam chamber into a copper coil immersed in $20 \text{ L}$ of water. The steam condenses in the coil and is returned to the steam chamber as water at $90^{\circ} \text{C}$. The latent heat of steam is $540 \text{ cal g}^{-1}$,and the specific heat of steam is $1 \text{ cal g}^{-1} {}^{\circ} \text{C}^{-1}$. The rise in temperature of the water is: (in $^{\circ} \text{C}$)