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, [tex]H_{supplied}[/tex] = 3.34 × 10³ J
The amount thermal energy gained by the kettle, [tex]H_{absorbed}[/tex] = 5.95 × 10² J
The percentage efficiency of the electrical element in heating the kettle of water, η%, is given as follows;
[tex]\eta \% = \dfrac{H_{supplied} - H_{absorbed} }{H_{supplied}} \times 100[/tex]
Therefore, we get;
[tex]\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 \%[/tex]
The percentage efficiency of the electrical element, η% ≈ 82.186%.
