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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Given:
- Power of the refrigerator = 750 W.
- Operates for 8 hours each day.
- Cost per kWh = Rs. 2.50.
To find: The cost of energy to operate it for the month of June.
Answer:
Power = 750 W
(To convert it to kW, divide the sum by 1000.)
Power = 750/1000 = 0.75 kWh
Energy = Power x Time.
Energy = 0.75 x 8
Energy = 6 Joules.
Total energy consumed in the month of June = 6 x 30 = 180 Joules.
(June has 30 days.)
Cost per kWh = Rs. 2.50.
∴ Cost for 180 kWh = 180 x 2.50 = 450
∴ The total cost is Rs. 450.