科學問題一問`請高手幫忙

電水爐功率4000w，操作時間為15min:

耗電量
= (每秒耗電量) x (秒數)
= (功率) x (秒數)
= (4000 W) x (15 x 60 s)
= (4000 J/s) x (15 x 60 s)
= 3600000 J

1 kWh = 3600000 J
耗電量
= 3600000/3600000 kWh
= 1 kWh

1 kWh 的電費是 $0.9
電費
= $0.9 x 1
= $0.9

**********
用相同方法計算：

燈功率是100W，操作時間為3小時。

耗電量
= (100 W) x (3 x 3600 s)
= 1080000 J

耗電量
= 1080000/3600000 kWh
= 0.3 kWh

電費
= $0.9 x 0.3
= $0.27

**********
電腦功率為300W，操作時間為30min:

耗電量
= (300 W) x (30 x 60 s)
= 540000 J

耗電量
= 540000/3600000 kWh
= 0.15 kWh

電費
= $0.9 x 0.15
= $0.135

E=能量
T=時間
功率=能量*時間
功率除以1000*3600
再用答案*每一度電價錢
(僅記把時間化成秒)
時間化成秒的方法
小時*60*60
分鐘*60

2009-01-31 00:33:26 補充：
*=乘

電水爐功率4000w，操作時間為15min，想找出耗電量(J)、耗電量(kWh)及電費。
4000 W = 4000/1000 kW = 4 kW
and 15 min = 15/60 hour
kWh (kilowatt hour) used by the stove
= 4 kW x 15/60 hr = 4x15/60 kWh = 4x15/60 x 3600000 J
Cost of electricity = $0.9 x 4x15/60

- 燈功率是100W，操作時間為3小時，想找出耗電量(J)、耗電量(kWh)及電費。
kWh used = (100/1000) W x 3 hr = 300/1000 kWh = 0.3 kWh
= 0.3 x 3600000 J
Electricity cost = $0.9 x 0.3 = $0.27

- 電腦功率為300W，操作時間為30min，想找出耗電量(J)、耗電量(kWh)及電費。
kWh used = (300/1000 kW x 30/60 hr = 9000/60000 kWh
= 9000/60000 x 3600000 J
Electricity cost = $0.9 x 9000/60000