A heater which consumes 800 Kilojoules/sec is on for 4hrs 15min. The energy consumed is
- 340 kWh.
- 4 kWh.
- 3400 kWh.
Explanation from PART66Online
The heater draws 800 kJ/s = 800 kW. Over 4 h 15 min = 4.25 h, energy = 800 kW * 4.25 h = 3400 kWh. Note the kilojoule-per-second rating makes this 1000 times larger than an 800 J/s load, so 3.4 kWh would be the answer only if the rating were 800 W.
A
arifadli8 asked · 8 Jan 2014
How?
Community Comments (1)
P
PART66Online
9 Jan 2014
kJ/s = kW.
800 * 4.25 = 3400 kWh.
800 * 4.25 = 3400 kWh.
Please Sign In to post a comment.
More discussions in 3. Electrical fundamentals
- A capacitor is fully charged after 25 seconds to a battery voltage of 20 Volts. The battery is replaced with a short circuit. What will be the voltage across the capacitor after one time constant?
- What must you do to make a 3 F capacitor circuit into a 2 F circuit?
- The time required to fully charge a 1 uF capacitor in series with a 100k ohm resistor is
- A 50 microFarad capacitor is fed with a current of 25 milliAmpere. How long will it take to charge it to 100Volts?
- A 50 microFarad capacitor is fed with a current of 25 milliAmpere. How long will it take to charge it to 100Volts?
- A 20 Picofarad capacitor with a 1 Megohm resistor takes how long to charge?