A 445g sample of ice at -58degrees is heated until its temp. reaches -28degrees. Find the change in heat content of the system.
thx
-xox-
2006-11-26 06:46:13 · 3 answers · asked by PinkFire007 2 in Science & Mathematics ➔ Chemistry
oops..the new temperature is -28degrees celcius
& the answer has to be in joules
forgot to mention that.
2006-11-26 06:53:50 · update #1
damn..again!
the new temperature is -29degrees..lol
2006-11-26 06:54:17 · update #2
The specific heat of ice is .5 cal/g-oC
--> http://www.school-for-champions.com/science/heat_ice_steam.htm
The ice changed temperature by 30 degrees
445 g * 30 degrees * .5 cal/g-oC =116.5 calories
2006-11-26 06:50:31 · answer #1 · answered by DanE 7 · 0⤊ 0⤋
Q = mc delta T
Q = (445 g)(specific heat of ice)(-28 - -58)
I don't remember the specific heat of ice. It will be in your text or just google it. -28 - -58 = 30 degrees C.
2006-11-26 06:50:15 · answer #2 · answered by physandchemteach 7 · 0⤊ 0⤋
Heat= ms(-28-(-58))
= 445(0.5)(30)
=6675 Cal. Ans.
2006-11-26 07:04:25 · answer #3 · answered by aminnyus 2 · 0⤊ 0⤋