- Joined
- Aug 24, 2005
- Messages
- 687
- Reaction score
- 4
- Points
- 4,591
Advertisement - Members don't see this ad
I didn't get this from any DAT study programs, but my chem. professor said problems like this sometimes appear on standardized tests with chemistry.
"A 55.4g sample of a metal was heated to 128 deg. C and then placed in a calorimeter containing 68.6 g of water at 20.3 deg. C. The final temperature of the water and metal in the calorimeter is 25.6 deg. C. Calculate the specific heat capacity of the metal. "
Here is what I have done so far:
heat lost by metal = heat gained by water
m X s X change in Temp. = m X s X change in Temp.
55.4g X s X 102.4 deg. C = 68.6g X 4.18 j/g-C X 5.3 deg. C
s = (68.6 g)(4.18 j/g-C)(5.3 deg. C) / (55.4)(102.4 deg. C)
s = 1519.7644/5672.96
This is where I'm not sure what's going on -- my math *should* be correct, but I'm not sure, since the top number is bigger than the bottom one.
What do you guys think?
"A 55.4g sample of a metal was heated to 128 deg. C and then placed in a calorimeter containing 68.6 g of water at 20.3 deg. C. The final temperature of the water and metal in the calorimeter is 25.6 deg. C. Calculate the specific heat capacity of the metal. "
Here is what I have done so far:
heat lost by metal = heat gained by water
m X s X change in Temp. = m X s X change in Temp.
55.4g X s X 102.4 deg. C = 68.6g X 4.18 j/g-C X 5.3 deg. C
s = (68.6 g)(4.18 j/g-C)(5.3 deg. C) / (55.4)(102.4 deg. C)
s = 1519.7644/5672.96
This is where I'm not sure what's going on -- my math *should* be correct, but I'm not sure, since the top number is bigger than the bottom one.
What do you guys think?
