Instructions:
1.Put contents into a pot containing 5 bowls of water.
2.After boiling, turn to mild flame and boil it until 3 bowls of water left.
How much energy at least is required to prepare the herbal tea according to the above instructions? You may assume that room temperature and mass of 1 bowl of water are 20 C and 150 g respectively.
(Given: Specific heat capacity of water = 4200 J kg–1 C–1;
Specific latent heat of vaporization of water = 2.26 106 J kg–1)
F.3物理問題
2007-08-09 8:56 pm
回答 (2)
2007-08-09 9:25 pm
✔ 最佳答案
The process involved 2 parts 1 , Temperature change from 20 C to 100 C
2 , 2 bowls of water is turned into steam
For part 1 , Temperature change related to
E = mc(temp change)
E = 0.15 x 4200 x (100-80)
E = 50400 (J)
For part 2, Change of state (liquid to gas) related to
E = ml
E = 0.15 x 2.26 x 10^6
E = 339000 (J)
In total , the least energy required is 339000+ 50400 = 389400 (J)
*Usually more energy is needed , because of heat loss to surrounding
*If an extra part require you an assumption , then it is to assume no heat loss to surrounding
*SI unit of mass is kg , so we express 150g to 0.15kg
2007-08-10 22:52:58 補充：
**謝樓下的提點 ! thxFor part 1 , 5 bowls of water 's Temperature change related to E = 5 x 0.15 x 4200 x (100-80)E = 63000 (J)For part 2, 2 bowls of water 's Change from liquid to gas, related to E = 2 x 0.15 x 2.26 x 10^6E = 678000 (J)In total 741000 (J)
2007-08-10 6:03 pm
上面的答案請將mass 乘2
收錄日期: 2021-04-13 01:00:56
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070809000051KK01932