A 745 g rectangular piece of Al at 285oC was dropped into 1.75 L of water at 27oC. What will be the temperature change of the metal? (Assume no heat is lost to the surroundings.)
A. 258oC
B. 49oC
C. 236oC
D. 21.6oC
C. 236 is the answer, but how do you get the answer?
I have two equations:
Heat= (Mass)(Specific heat)(Temp change)
Specific heat= (heat)/(mass x temp change)
Al specific heat is 0.900
Water specific heat is 4.184
Could someone show me how to get the answer
2007-06-19 05:21:32 · 4 answers · asked by kk a 1 in Science & Mathematics ➔ Chemistry
I don't know how to figure out how much heat in joules there is or how much the temperature will change.
2007-06-19 05:23:30 · update #1
C. 236oC
2007-06-19 05:27:51 · answer #1 · answered by ag_iitkgp 7 · 0⤊ 0⤋
calculate energy lost by the Al which is also the energy gained from water. formula:
-mcdelta t of Al=mcdelta t of water
delta t is the change in temperature, and c is the specific heat capacity for both. if you plug all the numbers in, you get the right answer.
2007-06-19 05:37:57 · answer #2 · answered by (^_^) CHiccaaqui(^_^) 4 · 0⤊ 0⤋
The energy lost from the metal was gained by the water.
I think B is the right answer.
-q(Al)=q(H2O)
-(754g)(.9)(Tf - 558) = (1750g)(4.184)(Tf - 300)
-678.6(Tf-558) = 7322(Tf-300)
-679Tf + 378324 = 7322Tf - 2196600
8001Tf = 2574924
Tf = 321.83 K
Tf = 49 C
2007-06-19 05:42:16 · answer #3 · answered by eric b 1 · 0⤊ 0⤋
B
2007-06-19 05:33:06 · answer #4 · answered by samantha 5 · 0⤊ 0⤋