Find the time required for an object to cool from 360°F to 250°F by evaluating the cooling formula t=(10/ln2)∫(the upper limit is 300 and the lower is 250) 1/(T-100)dT where t is time in minutes.
2006-12-11 00:54:55 · 5 answers · asked by Carter 2 in Science & Mathematics ➔ Mathematics
t=(10/ln2)∫(dT/(T-100) dT [Lims from 250 to 300]
t = (10/ln2) ( ln |T-100|) [Lims from 250 to 300]
t=(10/ln2) (ln|200| - ln|150|)
t=(10/ln2)(ln(200/150) = (10/ln2)ln(4/3)
2006-12-11 01:35:54 · answer #1 · answered by ironduke8159 7 · 0⤊ 0⤋
int. of 1/(T-100) dT = log of modulus of T-100
int. of 1/(T-100) dT from 250 to 300 = ln 200 - ln 150 =
ln (200/150) = ln (4/3) ( note: here base is "e" ) = 2ln 2 - ln 3 = .287682
t = 10/ln 2 (.287682) = 4.15037 minutes
using ( log (m/n) = log m - log n) & int of 1/T dT is log of mod of T
2006-12-11 01:06:59 · answer #2 · answered by Anonymous · 0⤊ 0⤋
t=10/ln2[lnT-100}limits
=10/ln2[{ln760-100}-{ln710-100}]
2006-12-11 01:03:33 · answer #3 · answered by raj 7 · 0⤊ 0⤋
Interested in this
2016-08-08 21:19:38 · answer #4 · answered by ? 3 · 0⤊ 0⤋
It depends..
2016-08-23 12:37:22 · answer #5 · answered by ? 4 · 0⤊ 0⤋