F.3 phy,,,about heat capacity!

2008-06-10 4:04 am
http://xbf.xanga.com/eedc607a11432193018195/w148818164.bmp

the most important thing is teaching me how to answer (d),,
i really dun know!!
i think it over and over again...still,,,dunno!!

for (d),,the answer is wrong,,
but why?!?
i think it should be true!

回答 (2)

2008-06-10 6:50 am
✔ 最佳答案
c)the temperature of the boiling is 100*c (normally)
so the metal block is also 100*c
let the specific heat capacity of metal is A*c^-1kg^-1 J
the water gain energy=the metal loss energy
0.3x4200x(38-23)=0.8xAx(100-38)
18900=49.6A
A=381*c^-1 kg^-1 J
the specific heat capacity of metal is 381*c^-1 kg^-1 J
d. How to measure specific heat capacity??
SHC=(the energy loss to the water)/(mass of the metal X the changing of temperature of the metal)
now,the mass and the changing of temperature is a constant..
then..if there is some energy lost to the surroundings..that means the energy that remain in the metal is lower...
if the energy that remain in the metal is lower..the energy loss to the water by the metal is also lower..
so the SHC will be lower not higher if the student is ture
so he is wrong
唔明歡迎再問~

2008-06-11 10:24:56 補充:
打得中文應該早d講 - -
why有temperature change??
就係因為兩者有temperature的差別先會有temperature change
有temperature的差別即係有能量的差別..
例如有兩杯水..兩杯都係1kg..一杯係2度...一杯係0度..
[[假設]]0度的水有0J的能量..假設咋!!(事實是要到攝氏-273.15度,才是0J的能量)
咁2度的1kg 就有2x4200x1=8400J
當兩者mix 埋..能量就互相分均..到最後溫度相等..
最後溫度是8400/2/4200=1度..

2008-06-11 10:28:56 補充:
但假設2度那杯水..LOST 咗4200j to surrounding..
咁就得返1度啦..再mix0度個杯..
咁就得返4200/2/4200=0.5度..
而loss 去0度杯水就有2100j(比4200細咗)..
wow..原來the changing of temperature of the 2度的水都會增大(由1度-->1.5度)
睇返shc條式..
咁依家分子細咗..分母又大咗..咁個shc就會更細..
令shc變大的可能就是kewym99所說的..metal帶埋少少熱水過去..
2008-06-10 6:49 am
The explanation was wrong for sure. Consider the situation:
When the metal block was taken out from the hot water, assuming heat loss to the surroundings, made the metal block cooler than it was in the beaker. Eventually the final temperature of water in the cup measured was lower than expected and the calculated specific heat capacity was lower than the true value, not higher.
There are other possible explanations for a specific heat capacity higher than the true value, for example, the metal block was not dry during the transfer and some hot water was transferred to the cup.


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