Question

If an electric kettle of power \(1.5\,\text{kW}\) connected to a \(220\,\text{V}\) supply increases the temperature of water of mass \(1600\,\text{g}\) from \(20^\circ\text{C}\) to \(95^\circ\text{C}\) in \(7\) minutes, then the efficiency of the kettle is \[ \left(\text{Mechanical equivalent of heat}=4.2\,\text{J cal}^{-1}\right) \]

• A. \(60\%\)
• B. \(75\%\)
• C. \(80\%\)
• D. \(90\%\)

Hint:

Efficiency is \[ \boxed{ \eta=\frac{\text{Useful output energy}}{\text{Input energy}}\times100\%. } \] For heating, \[ \boxed{ Q=mc\Delta T. } \]

Answer 1

The Correct Option is C

Step 1: Calculate the useful heat gained by water. \[ m=1600\,\text{g}, \qquad \Delta T=95-20=75^\circ\text{C}. \] Heat gained, \[ Q = mc\Delta T = 1600\times75 = 1.2\times10^5\ \text{cal}. \] Converting into joules, \[ Q = 1.2\times10^5\times4.2 = 5.04\times10^5\ \text{J}. \]

Step 2: Calculate the electrical energy supplied. Power, \[ P=1.5\,\text{kW}=1500\,\text{W}. \] Time, \[ t=7\times60=420\,\text{s}. \] Electrical energy supplied, \[ E=Pt =1500\times420 =6.3\times10^5\ \text{J}. \]

Step 3: Find the efficiency. \[ \eta = \frac{Q}{E}\times100 = \frac{5.04\times10^5}{6.3\times10^5}\times100 = 80\%. \] Hence, \[ \boxed{80\%} \] Therefore, \[ \boxed{(C)} \] is the correct answer.