J.R. S. answered • 04/15/20
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Look up the specific heat for platinum. I find it to be 0.13 J/g/degree
q = mC∆T
q = heat = ?
m = mass = 14.9 g
C = specific heat = 0.13 J/g/degree
∆T = change in temperature = 40.0 - 20.8 = 19.2 degrees
q = (14.9 g)(0.13 J/g/degree)(19.2 degrees)
q = 37.2 J
J.R. S.
tutor
Well, you didn't provide a value for the specific heat for platinum, and if they used a different value from 0.13, then the answer will be different. The process is correct, you just need to use the correct values, and do the math correctly.
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04/15/20
Amberly T.
My hw said the answer was wrong04/15/20