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Electric water heaters have heating elements mounted in the tank. The input current in the water...

Electric water heaters have heating elements mounted in the tank. The input current in the water heater is 3.56 A and potential difference (Voltage) across it is 240 V. When the heater is switched on for 10 minutes, what could be the input power and electrical energy (in kWh) taken by water heater? If the efficiency of water heater is found to be 90 % due to energy loss, then how will you evaluate the practical output power generated by the heater to heat up water. Can you identify two reasons for this loss of energy since efficiency is 90 %?
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Answer #1

A)Power = Voltage* current

P=240*3.56

P= 854.4 watt

B)Electrical energy = power*time

E= 854.4*600 = 512,640 joule = 0.1424 Kwh

C) we will measure the rise in temperature and mass of water and we already know the specific heat capacity of water

We will calculate Q and compare with Ideally calculated values or just find ratio of them from here we will get idea of power

D) (1) heat lost to environment because heater body is not perfectly adiabatic so that no heat can escape from walls

(2) some energy is Lost in the form of eddy currents and hysteresis loss

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