Question

Two litres of water at initial temperature of ${27}^{\circ }\text{C}$ is heated by a heater of power $1\text{kW}$ in a kettle. If the lid of the kettle is open, then heat energy is lost at a constant rate of $160\text{J/s}$. The time in which the temperature will rise from ${27}^{\circ }\text{C}$ to ${77}^{\circ }\text{C}$ is (specific heat of water $=4.2\text{kJ/kg}$).

• A. $5\text{min}20\text{sec.}$
• B. $8\text{min}20\text{sec.}$
• C. $10\text{min}40\text{sec.}$
• D. $8\text{min}50\text{sec.}$

Answer 1

The correct option is B $8\text{min}20\text{sec.}$

From principle of calorimetry,
Heat gained by water = Heat supplied - Heat loss
$ms\Delta \theta =1000t-160t\Rightarrow t=\frac{2\times 4200\times 50}{840}=8\text{min.}20\text{sec.}$