• question_answer Two litres of water of at initial temperature of ${{27}^{o}}C$ is heated by a heater of power 1KW in a kettle. If the lid of the kettle is open, then heat energy is lost at a constant rate of$160J/s$. The time in which the temperature will rise from ${{27}^{o}}C$ to $77{}^\circ C$ is (specific heat of water$=4.2\text{ }KJ/kg$): A)  $5\text{ }min\text{ }20\text{ }s$                   B)  $8\text{ }min\text{ }20\text{ }s$ C)  $10\text{ }min\text{ }40\text{ }s$                 D)  $12\text{ }min\text{ }50\text{ }s$

As, heat gained by water = Heat supplied - heat loss $ms\Delta \theta =1000t-100t$(by the question)             $t=\frac{2\times 4200\times 50}{840}=8\,\min \,20s$