2 litre water at  $27°C$ is heated by a 1 kW heater in an open container. On an average heat is lost to surroundings at the rate of 160 J/s. The time required for the temperature to reach $77°C$  is

From question heat gained by water in $1\text{second}=1KW-160J/S=1000160J/s=840J/s$ 

Total heat required to raise the temperature of water(volume $2\mathrm{L})$ from 27 to ${77}^{0}C$

$=m\left(water\right)\times sp.ht\times \Delta \theta$

$=2\times 10\wedge 3\times 4.2\times 50$ $[Q\text{mass}=\text{density}\times \text{volume}]$

Thus 

$840\times t=2\times 10\wedge 3\times 4.2\times 50$

OR
$\begin{array}{l}t=2\times 10\wedge 3\times 4.2\times 50/840\\ t=500\sec onds\\ t=8\min 20\sec \end{array}$