# Help with a basic physics questionWatch

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Thread starter 2 years ago
#1
An old fridge works for 24 hours a day, every day of the year.
It has a current of 0.63 A from the 230 V mains.
(a) Which of the following values is the energy in kilowatt hours transferred in a year, written to two significant figures?
Assume that a year is 9000 hours.
Put a ring around the correct value in kWh.
1300 , 3200 , 1 300 000 , 3 200 000

So, surely you work out the power in Watts first:
- 0.63A * 230V = 144.9W = 144.9 J / s

Now, covert it into Watt Hours:
- 144.9W * 60 * 60 = 521,640Wh

Now, work it out for the year (which is roughly 9000 hours):
- 521, 640Wh * 9000 = 4,694,760,000Wh in a year

Covert it into kWh:
- 4,694,760,000Wh / 1000 = 4,694,760kWh = 4,700,000 (2 d.p.)

The mark scheme does't covert the power in watts into watt hours, and gets a value of 1300, but I don't see why and where I have gone wrong. Any help would be greatly appreciated! Thanks.
0
Thread starter 2 years ago
#2
bump - My exam's today
0
2 years ago
#3
(Original post by thomas-cowley)
An old fridge works for 24 hours a day, every day of the year.
It has a current of 0.63 A from the 230 V mains.
(a) Which of the following values is the energy in kilowatt hours transferred in a year, written to two significant figures?
Assume that a year is 9000 hours.
Put a ring around the correct value in kWh.
1300 , 3200 , 1 300 000 , 3 200 000

So, surely you work out the power in Watts first:
- 0.63A * 230V = 144.9W = 144.9 J / s

Now, covert it into Watt Hours:
- 144.9W * 60 * 60 = 521,640Wh

Now, work it out for the year (which is roughly 9000 hours):
- 521, 640Wh * 9000 = 4,694,760,000Wh in a year

Covert it into kWh:
- 4,694,760,000Wh / 1000 = 4,694,760kWh = 4,700,000 (2 d.p.)

The mark scheme does't covert the power in watts into watt hours, and gets a value of 1300, but I don't see why and where I have gone wrong. Any help would be greatly appreciated! Thanks.
Yeah, you didnt convert to WattHours, you converted to Joules/hour... to convert to watt hours multiply the power (in watts) by the time (in hours).
1
2 years ago
#4
(Original post by thomas-cowley)
An old fridge works for 24 hours a day, every day of the year.
It has a current of 0.63 A from the 230 V mains.
(a) Which of the following values is the energy in kilowatt hours transferred in a year, written to two significant figures?
Assume that a year is 9000 hours.
Put a ring around the correct value in kWh.
1300 , 3200 , 1 300 000 , 3 200 000

So, surely you work out the power in Watts first:
- 0.63A * 230V = 144.9W = 144.9 J / s

Now, covert it into Watt Hours:
- 144.9W * 60 * 60 = 521,640Wh

Now, work it out for the year (which is roughly 9000 hours):
- 521, 640Wh * 9000 = 4,694,760,000Wh in a year

Covert it into kWh:
- 4,694,760,000Wh / 1000 = 4,694,760kWh = 4,700,000 (2 d.p.)

The mark scheme does't covert the power in watts into watt hours, and gets a value of 1300, but I don't see why and where I have gone wrong. Any help would be greatly appreciated! Thanks.
Power in watts = 230 x 0.63 = 144.9W
Since they want it in kWh, divide by 1000 to get the answer in kWh.
Then multiply your answer by 9000 to get the kWh in a year
So the answer of 1300 is correct.
Hope that helps
1
Thread starter 2 years ago
#5
Thanks everyone! This really helped as a similar question came up in the exam.
0
2 years ago
#6
(Original post by thomas-cowley)
Thanks everyone! This really helped as a similar question came up in the exam.
No worries! What board did you do? I did OCR 21st century and a similar question came up for us too.
0
Thread starter 2 years ago
#7
(Original post by kennethdcharles)
No worries! What board did you do? I did OCR 21st century and a similar question came up for us too.
Yeah that as the same board as me, how did you find it?
0
2 years ago
#8
It was harder than every other paper... but nevertheless I still found it pretty decent, and should be happy with my result You?

(Original post by thomas-cowley)
Yeah that as the same board as me, how did you find it?
0
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