有時睇一條應用題計heat個d,我地點可以睇到個個carrier有冇involve 到個temperature change a?
eg: a student puts 0.4kg of water into a iron container of heat capacity 120j/degree . A heater of power 200 W is used to raise the temperature of the water from20 degree to 80 degree.
ai} assume that there is no heat loss to the surroundings air, what are energy required?
點解卓個energy gain個時要計埋個iron container ge energy gain ge?
唔該曬=]
應用題計heat ge 一問{唔該多d高手黎答..}
2007-02-11 6:36 am
回答 (2)
2007-02-12 4:31 am
✔ 最佳答案
因為加熱...係先加去個iron container~再傳比d水~有熱傳播既條件係高溫去低溫~直至兩個溫度一樣~
即係話個container唔會比哂d熱d水~
都冇可能話唔計個container...如果佢有比到資料你...就要計啦~
佢問題話將水加到80度~
即係話個iron container都大慨起碼去到80度~
因為container冇咁熱...水又點有咁熱~
所以你個answer要計兩樣野~
水吸左既熱+container吸左既熱~
水既specific heat capacity=4200J/(kg K)
水gain既energy=[4200x0.4x(80-20)]=100800j
iron container gain既energy=120 x (80-20)=7200j
total energy required=7200+100800=108000J
如果問埋用幾多時間既話,108000/200=540s
參考: 自己
2007-02-11 6:44 am
0的水裝在個iron container ﹐題目說將水由20 度 加到80 度.
但係你要考慮你煲水是隔住個iron container 來煲的﹐所以點樣講都是個煲吸收了熱能再傳給0的水吧。
若果將水加熱到80度﹐則個煲大概也會被加到80度
所以energy required包括
將水加熱到80度所需的能量+將煲加熱到80度所需的能量
2007-02-10 22:47:09 補充：
名詞時的煲指iron container
但係你要考慮你煲水是隔住個iron container 來煲的﹐所以點樣講都是個煲吸收了熱能再傳給0的水吧。
若果將水加熱到80度﹐則個煲大概也會被加到80度
所以energy required包括
將水加熱到80度所需的能量+將煲加熱到80度所需的能量
2007-02-10 22:47:09 補充：
名詞時的煲指iron container
收錄日期: 2021-04-25 16:53:28
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20070210000051KK04895