Question

in heating a kettle of water on an electric stove, 3.34×10^3 J of thermal energy was provided by the element of the stove. yet, the water in the kettle gained only 5.95×10^2 J of thermal energy. determine the percent efficiency of the electrical element in heating the kettle of water​

Answer 1

The percentage efficiency of the electrical element is approximately 82.186%

Step-by-step explanation:

The given parameters are;

The thermal energy provided by the stove element,
`H_(supplied)` = 3.34 × 10³ J

The amount thermal energy gained by the kettle,
`H_(absorbed)` = 5.95 × 10² J

The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;

`\eta \% = (H_(supplied) - H_(absorbed) )/(H_(supplied)) * 100`

Therefore, we get;

`\eta \% = (3.34 * 10^3 - 5.95 * 10^2)/(3.34 * 10^3) * 100 = (549)/(668) * 100 \approx 82.186 \%`

The percentage efficiency of the electrical element, η% ≈ 82.186%.