118k views
0 votes
an electrical kettle has a power rating of 100 W. it takes 4 minutes to boil the water. if it takes 1.96 x 10^(5) [196 000J] of energy to boil the water, what is the efficiency of the kettle?

User Qwe Asd
by
8.2k points

1 Answer

5 votes

Final answer:

The efficiency calculation results in an implausible value of 816.67%, which is impossible as efficiency cannot exceed 100%. This suggests there is an error in the provided data or calculations.

Step-by-step explanation:

To calculate the efficiency of the kettle, we need to use the formula efficiency = (useful energy output / total energy input) × 100%. Here, the useful energy output is the energy required to boil the water, which is given as 1.96 × 105 J (196,000 J), and the total energy input can be calculated from the power rating of the kettle and the time it took to boil the water. The total energy input is power × time = 100 W × 4 × 60 s = 24,000 J since power is in watts (Joules per second) and time is in minutes, we convert time to seconds. The efficiency can then be calculated as follows:

(196,000 J / 24,000 J) × 100% = 816.67%

However, it is impossible for any appliance to have an efficiency greater than 100%, which indicates that there is either a mistake in the power rating, the energy required, or the time given. We must recheck the data or the calculations for accuracy because there may be an error present.

In the context given, if the kettle could heat water with 100% efficiency, the 100 W kettle would take much longer to boil the water since only 24,000 J of energy is supplied in 4 minutes, which is significantly less than the required 196,000 J to boil the water.

User Govind Totla
by
8.4k points