In heating a kettle of water on an electric stove, 3.34 x 10^4 J of thermal energy was provided by the element?
In heating a kettle of water on an electric stove, the heating element provided 3.34 x 10^4 J of thermal energy. However, the water in the kettle only absorbed 5.95 x 10^2 J of thermal energy. Calculate the efficiency percentage of the electric element in heating the kettle of water. Please show your work in the calculations.
3 Answers
Divide the energy gained by the water by the energy produced by the stove element. That will will be the fractional efficiency. Multiply by 100% to convert that to percentage.
5.95*10^2J / 3.34*10^4J = 1.78143 x 10^-2 = 0.0178 fractional efficiency
0.0178 * 100% = 1.78%
This process had an efficiency of 1.78%, while the rest of the energy was lost to heating the pot itself, as well as the atmosphere (by convection) and the walls and ceiling (by radiation).
Even the right answers don’t have the right reason… Kettle on gas stove. You avoid the conversion from fuel to electricity at the powerplant, and the loss in conversion from electricity to microwaves (around 50%). But this only applies if you don’t heat a lot more water than you’re about to use.
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