在浴室內,浴缸的冷熱水龍頭同時打開3分鐘。
熱水在通過熱水器後流出的速率是0.3kgs-1,溫度為70oC。而於水流出的速率是0.2kgs-1,溫度為20oC。
(a)假設能量沒有在管道中散失,求熱水器供應能源的速率。
(b)在浴缸內,水的最後溫度是多少?
(c)當浸浴前,我們會先用手攪拌浴缸內的水。試解釋這個動作。
物理 - 熱學
2009-04-12 6:20 am
回答 (2)
2009-04-12 9:42 am
✔ 最佳答案
(a) Consider a time interval of 1 s,heat absorbed by water = 0.3 x 4200 x (70-20) J = 630 000 J
where 4200 J/kg-K is the specific heat capacity of water
hence, power of heater = 630 kW
(b) Amount of heat given by hot water
= 0.3 x (3x60) x 4200 x (70-T) J
where T is the final temperature of the mixture
Amount of heat absorbed by cold water
= 0.2 x (3x60) x 4300 x (T-20) J
using the heat balance equation,
0.3 x (3x60) x 4200 x (70-T) = 0.2 x (3x60) x 4300 x (T-20)
solve for T
(c) To well mix the hot and cold water in order to obtain a uniform water temperature.
2009-04-12 6:35 am
只係識答
C)用手攪拌浴缸內的水係為左令水既溫度更平均!
C)用手攪拌浴缸內的水係為左令水既溫度更平均!
收錄日期: 2021-04-23 20:39:05
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