物理問題 明天要交
1.那個雪糕質量為30kg的男孩吃下0.3kg的雪糕(0℃),假設雪糕皈冰兩者之間的比潛熱大約相同,即係334000jkg-1℃
(a)計算該名男孩在吃下雪糕後的體溫.
2.用火把0.2kg的鋁塊加熱,然後迅速把它放進1kg的水中,水的溫度為20℃.由聚苯乙烯杯盛載.若水的末溫度是60℃.計算火燄的溫度.(鋁的比熱容量為900jkg-1℃-1)[答案是993℃]
3.電熱器的輸出功率為200w.液體沸騰300s後.秤的讀數下降了15g.如果散失到周圍環境中的能量可以忽略不計,那麽,液體的汽化比潛熱是多少?
(答案是4x10^6jkg-1)
物理問題 明天要交
2006-11-29 5:07 am
回答 (2)
2006-12-01 3:47 am
✔ 最佳答案
1.) ( 先計雪糕由固體變液體所需之能量)E=mL_______________________E=energy,m=mass,L=latent heat
E=(0.3)(334000)
E=100200
(然後計雪糕變液體時令人改變多少度)
E=mc△T___E=energy,m=mass,c=specific heat capicty,△T=change in temperature
100200=(30)(4200)(△T)
△T=0.7952
(係度我假設雪糕的比熱容量同水一樣,因為你無比到data,如果唔一樣就自己改個data,而人的溫度就係37度減去將雪糕由固體變液體所改變了的溫度)
m1c1(T1-T)=m2c2(T-T2)___________m=mass,c=specific heatcpacity,T=temperature
(30)(4200)[(37-0.7952)-T]=(0.3)(4200)(T-0)
(4561804.8)-(126000T)=1260T
T=35.846=35.8℃
2.) m1c1(T1-T)=m2c2(T-T2)___________m=mass,c=specific heatcpacity,T=temperature
(0.2)(900)(T1-60)=(1)(4200)(60-20)
180T1-10800=168000
T1=993.333=993℃
3.) E=Pt________________________E=energy,P=power,t=time
E=(200)(300)
E=60000
E=mL_______________________E=energy,m=mass,L=latent heat
(60000)=(0.015)(L)
L=4000000=4x10^6jkg-1
4.) E=Pt________________________E=energy,P=power,t=time
E=mc△T___E=energy,m=mass,c=specific heat capicty,△T=change in temperature
Pt=mc△T
P(10x60)=m(4200)(100-15)
P=m(4200)(100-15)/(10x60)
E=mL_______________________E=energy,m=mass,L=latent heat
Pt=mL
Pt=m(2.26x10^6)
P=m(2.26x10^6)/t
m(4200)(100-15)/(10x60)=m(2.26x10^6)/t
(4200)(100-15)/(10x60)=(2.26x10^6)/t
t=3798.319s=63.305mins=63.3mins
2006-11-29 5:40 am
1.資料唔充足
2.0.2x900(t-60)=1x4200x(60-20)
180t-10800=16800
t=993.33333
t=993c(corr.to nearest c)
3.E=PT
E=200X300
E=60000
E=MLv
60000=1500XLv
Lv=4000000
=4x10^6jkg-1
2.0.2x900(t-60)=1x4200x(60-20)
180t-10800=16800
t=993.33333
t=993c(corr.to nearest c)
3.E=PT
E=200X300
E=60000
E=MLv
60000=1500XLv
Lv=4000000
=4x10^6jkg-1
收錄日期: 2021-04-13 16:10:04
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20061128000051KK04196