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

$A$ thermos flask contains $250 \ g$ of coffee at $90^{\circ} C$. To this $20 \ g$ of milk at $5^{\circ} C$ is added. After equilibrium is established, the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 \ cal/g^{\circ} C$) (in $^{\circ} C$)

Three containers $C_{1}, C_{2}$ and $C_{3}$ have water at different temperatures. The table below shows the final temperature $T$ when different amounts of water (given in litres) are taken from each container and mixed (assume no loss of heat during the process).

$C_{1}$	$C_{2}$	$C_{3}$	$T$
$1 \ l$	$2 \ l$	$-$	$60^{\circ} C$
$-$	$1 \ l$	$2 \ l$	$30^{\circ} C$
$2 \ l$	$-$	$1 \ l$	$60^{\circ} C$
$1 \ l$	$1 \ l$	$1 \ l$	$\theta$

The value of $\theta$ (in $^{\circ} C$ to the nearest integer) is

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

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

An electric kettle (rated at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\, ^oC$ to its boiling point. It takes $9.5$ minutes. The amount of heat lost is: