Answer to Question #119552 in Molecular Physics | Thermodynamics for Sadie M

Question

A kettle transfers 6000 J of energy electrically. 1500 J of this is wasted. What is the efficiency of this kettle?

Answer 1

We can find the efficiency of the kettle from the formula:

$$Efficiency = \dfrac{useful \ power \ transferred}{total \ power \ supplied} \cdot 100\%.$$

The useful power transferred by the kettle can be found from the formula:

$$Useful \ power \ transferred = Supplied \ electrical \ energy - wasted \ energy,$$ $$Useful \ power \ transferred = 6000J - 1500J = 4500J.$$

Then, we can calculate the efficiency of the kettle:

$$Efficiency = \dfrac{4500J}{6000J} \cdot 100\% = 75\%.$$

Answer:

$Efficiency = 75\%.$