一部自動洗衣機內有3000W的發熱器,把水升至所需的溫度需時14分鐘。
(a)發熱器在14分鐘內供應了多少能量(840秒)？
(b)洗衣機用水20kg,水的比熱容量為4200Jkg-1 oC-1。
(i)利用(a)項的答案,求水溫上升的幅度。
(ii)為甚麼實際上,水溫的上升幅度比上項結果略小？
(c)這洗衣機洗少量衣服時,可調校輕量掣,使其用一半的水量運作。
(i)把10kg的水升高至相等的溫度,需要多少能量？
(ii)為甚麼把衣服用一次全負荷運作,比較把衣服分為兩次用輕量運作更為經濟有效？
physics物理~~~~~~~~~~~~
2010-01-20 7:21 am
回答 (2)
2010-01-22 5:58 am
✔ 最佳答案
(a) By Energy = Power x Time (E = Pt)E = 3000W x 840s = 2520000 J
(b) (i) By Energy = mass x specific heat capacity x temperature change
(E = mc delta T),
2520000 = (20)(4200)(delta T)
delta T = 30 C
(ii) 因為能量流失在周邊空間 (energy is lost to the surroundings), 加上部分能量需令到發熱器變熱,致實際上加熱水的能量比計算出來的能量少,By E = mc (delta T),能量和水溫轉變成正比,令到水溫的上升幅度比上項結果略小。
(c) (i) By E = mc delta T,
E = (10)(4200)(30)
= 1260000J
(ii) 2次輕量運作需要的能量 = 1260000 x 2 = 2520000J
1次全負荷運作需要的能量 = 2520000J
2次輕量運作需要的能量 = 1次全負荷運作需要的能量
基於能量相同,但兩次用輕量運作比1次全負荷作需要的時間少,
故把衣服用一次全負荷運作,比較把衣服分為兩次用輕量運作更為經濟有效。
2010-01-20 5:04 pm
(a) Heat delivered = 3000 x (14 x60) J = 2 520 000 J
(b) (i) Temperature rise = 2520000/(20x4200) 'C = 30'C
(ii) There is heat lost to the surroundings.
(c) (i) Heat required = 10 x 4200 x 30 J = 1 260 000 J
(ii) Because there is heat lost to the surroundings each time the washing machine is in operation, washing of clothes in two separate time means doubling the heat loss, thus leading to to waste of energy.
(b) (i) Temperature rise = 2520000/(20x4200) 'C = 30'C
(ii) There is heat lost to the surroundings.
(c) (i) Heat required = 10 x 4200 x 30 J = 1 260 000 J
(ii) Because there is heat lost to the surroundings each time the washing machine is in operation, washing of clothes in two separate time means doubling the heat loss, thus leading to to waste of energy.
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