$A$ $210 \,W$ 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 $\,s$)

Three copper blocks of masses $M_1, M_2$ and $M_3$ $kg$ respectively are brought into thermal contact until they reach equilibrium. Before contact,they were at temperatures $T_1, T_2, T_3$ $(T_1 > T_3)$. Assuming there is no heat loss to the surroundings,the equilibrium temperature $T$ is ($s$ is the specific heat of copper).

If an electric heater is rated at $1000\, W$, then the time required to heat one litre of water from $20\, ^oC$ to $60\, ^oC$ is

$A$ tap supplies water at $10\,^{\circ}C$ and another tap at $100\,^{\circ}C$. How many $kg$ of hot water must be taken so that we get $20\, kg$ of water at $35\,^{\circ}C$?

In an industrial process,$10 \, kg$ of water per hour is to be heated from $20^\circ C$ to $80^\circ C$. To do this,steam at $150^\circ C$ is passed from a boiler into a copper coil immersed in water. The steam condenses in the coil and is returned to the boiler as water at $90^\circ C$. How many $kg$ of steam is required per hour? (Specific heat of steam $= 1 \, cal/g^\circ C$,Latent heat of vaporisation $= 540 \, cal/g$)

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