Question

What must be the efficiency of an electric kettle marked \[500W,230V\] if the was found to bring \[1kg\] of water at \[15^\circ C\] to boiling point in \[15\min s\] ? (given specific heat capacity of water= \[4200J/kg^\circ C\] )
(A) \[79\% \]
(B) \[81\% \]
(C) \[72\% \]
(D) \[69\% \]

Answer 1

Hint: We are asked to find the efficiency of the kettle given in the question. For this we can start by defining efficiency. We can define efficiency (in this case) as the ratio of heat absorbed to the heat produced. Initially we find the two values and get the ratio but efficiency is expressed in terms of percentage. So, we multiply the value we get with a hundred percent thus we get the required solution.

Formula used: The formula used to find the heat produced is \[{Q_{produced}} = VIt\]
The formula used to find the heat absorbed is \[{Q_{absorbed}} = mC\Delta T\]
The efficiency of the electric kettle is given by the formula \[\eta = \dfrac{{{Q_{absorbed}}}}{{{Q_{produced}}}} \times 100\% \]
Where \[V\] is the voltage
\[I\] is the current
\[t\] is the time
\[m\] is the mass of substance
\[\Delta T\] is the change in temperature
\[C\] is the specific heat of water


Complete step by step solution:
Let us start by noting down the values given
The power of the kettle is \[P = VI = 500\]
The time taken is \[t = 15 \times 60 = 900s\]
The mass of the water is \[m = 1kg\]
The specific heat capacity of water is \[C = 4200J/kg^\circ C\]
The difference in temperature is \[\Delta T = 100 - 15 = 85\]
We can find the heat produced using the formula \[{Q_{produced}} = VIt\]
Substituting we get \[{Q_{produced}} = VIt = 500 \times 15 \times 60 = 4.5 \times {10^5}J\]
Now that we have found the heat produced, we can move onto finding the heat absorbed
The heat absorbed is given by the formula \[{Q_{absorbed}} = mC\Delta T\]
Substituting we get \[{Q_{absorbed}} = mC\Delta T = 1 \times 4200 \times 85 = 357 \times {10^3}\]
We can divide the two and get efficiency; that is, \[\eta = \dfrac{{{Q_{absorbed}}}}{{{Q_{produced}}}} \times 100\% \]
Substituting we get \[\eta = \dfrac{{{Q_{absorbed}}}}{{{Q_{produced}}}} \times 100\% = \dfrac{{357 \times {{10}^3}}}{{4.5 \times {{10}^5}J}} \times 100\% = 79.33\% \]
In conclusion, the right answer is option (A) \[79\% \]

Note:
Efficiency gives us the highest performance using the least quantity of input that gives us the highest output. The concept of efficiency plays a huge role in the energy crisis. Efficiency is important because it helps to reduce unnecessary resources like time and manpower which are essential for many other purposes.