Question

In heating a pot of water, 2.00 x 10^3J of heat was supplied by the stove element. If only 5.00 x 10^2J of heat was actually gained by the water (heat output), what was the percent efficiency of stove element?

Answer 1

The percent efficiency of the stove element is 25%.

Step-by-step explanation:

To find the percent efficiency of the stove element, we need to calculate the efficiency using the formula:

Efficiency = (useful output / input) x 100%

In this case, the useful output is the heat gained by the water, which is 5.00 x 10^2J, and the input is the heat supplied by the stove element, which is 2.00 x 10^3J.

Using the formula: Efficiency = (5.00 x 10^2J / 2.00 x 10^3J) x 100%

Simplifying, we get:

Efficiency = 25%