An electric refrigerator rated as 750 W operates 8 hours per day. What is the cost of the energy to operate it for a month of June at Rs. 2.50 per kWh?
Answers
Answer:
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Explanation:
Given, Power, P = 750 W
Time, T = 8 hrs per day
We know that, electrical energy is the product of power and time.
Therefore, the energy consumed by the refrigerator in one day
= 750 W × 8.0 hour/day
= 6000 W h = 6 kW h
Now, the total energy consumed by the refrigerator in the month of June (30 Days)
= 30 × Energy/day
= 30 × 6kWh
=180 kWh
Thus, the cost of energy required to operate the refrigerator for the month of June at Rs. 2.50 per kWh = 180 kW h × Rs 2.50 = Rs 450.00 !
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Hi !
Power = P = 750 W = 750/1000 kW
(1 kW = 1000 W ) , ( 1 W = 1/1000 kW)
Time = t = 8 hrs
Energy = P*t = 750/1000*8 = 0.75*8 = 6 J
Energy consumption in one day = 6 J
Energy consumption in the month of June = 6*30 = 180 J
Total cost = monthly consumption*rate
= 180*2.50
= Rs 450
The answer is Rs 450
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