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

Answer:

The percentage efficiency of the electrical element is approximately 82.186%

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;

$\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}} \times 100$

Therefore, we get;

$\eta \% = \dfrac{3.34 \times 10^3 - 5.95 \times 10^2}{3.34 \times 10^3} \times 100 = \dfrac{549}{668} \times 100 \approx 82.186 \%$

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