Suppose that you take a thermometer outside where it is 100°.(adsbygoogle = window.adsbygoogle || []).push({});

T(5min)=80° T(15min)=90°

What is the initial temp of the thermometer?

Given equation

dT/d t= k(T-Te)

Derived Equation

⌠(T-Te)^-1 (dT/dt)dt =⌠ kdt

ln(T-Te)=kt + c

T=ce^kt + Te

so i basically got the answer by knowing c must be negative and that when i use a square root on the magnitude of c i through out the positive value.

Wolfram Alpha

my real question is how can this become -3 ln(-20/x) +ln(-10/x) becomes -2ln(-(1/x)) -ln(800)?

Attempt to find C

1) 80=ce^k5 +100 ---- -20=ce^k5 ----- -20/c=e^k5 ------ ln(-20/c)=k5 ----- -3ln(-20/c)=-k15

2) 90=ce^k15 +100 ---- -10=ce^k15 ----- -10/c=e^k15 ------ ln(-10/c)=k15

becomes

ln(-10/c) - 3ln(-20/c) ----- ln((-10/c) x (c^3 / -8000)) -------- ln(c^2/800) ---- 2ln(c) -ln(800)=0

ln(c)=(1/2)ln(800)

c= +/- sqrt(800)

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# I Find This Initial Temperature

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