Physics easy question
2008-01-13 6:19 pm
A 1kW heater, immersed in water (0.5kg , 20°C )is switched on for 10 min .Calculate the maximum amount of water boiled away.
回答 (1)
2008-01-13 7:19 pm
✔ 最佳答案
total energy provide:P= E/T
1000 = E/(10X60)
E = 1000X600
= 600000J
let the maximum amount of water boiled away be y:
0.5(4200)x(100-20) + y(2.26X10^6) = 600000
168000 + y(2260000) = 600000
y(2260000) = 432000
y = 0.191
maximum amount of water boiled away is 0.19kg
參考: ME
收錄日期: 2021-04-19 00:16:34
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