Anonymous

# Newton's Law of Cooling?

so since Newton's Law says, T'(t) = k (C - T(t))

and its for a pot of boiling water removed from heat to on table in a room

C is room temperature

T(0) = 100 and it cools down to 84C in 2 mins and after another 2 mins, its down to 72C

how would i find T(t) equation & find room temperature???

Relevance

dT/dt = k (C - T)

∫ dT / (C - T) = ∫ k dt

- ln |C - T| = kt + c

ln |C - T| = -kt + c

e^(ln |C - T|) = e^(-kt + c) = e^(-kt) * e^(c)

C - T = Ae^(-kt)

where A = ± e^(c)

T(t)= Ae^(-kt) + C

since C is room temp, C = ~24˚ C

so now plug in given values...

T(0) = Ae^(-k(0)) + 24 = 100

A = 76

T(t) = 76e^(-k(t)) + 24

T(2) = 76e^(-k(2)) + 24 = 84

k = - ln[(84-24)/76] / 2 = ~ 0.11819

therefore:

A(t) = 76e^(-(0.11819)(t)) + 24

This is an approximate equation, as I don't know exactly what is considered room temperature. However, the method to solve this type of problem should now be apparent.

You can find the temperature of the room by just leaving C and A in the equation for a substitution method...

T(t)= Ae^(-kt) + C

T(0)= Ae^(-k(0)) + C = 100

A + C = 100

C = -A + 100

T(2)= Ae^(-k(2)) + (-A + 100) = 84

Ae^(-k(2)) - A = -16

A (e^(k(2)) - 1) = -16

(e^(k(2))) = (-16 / A) + 1

k = ln[ (-16 / A) + 1 ] / 2

T(4)= Ae^([ln[ (-16 / A) + 1 ] / 2](4)) + (-A + 100) = 72

Ae^(ln[ (-16 / A) + 1 ](2))+ (-A + 100) = 72

A (-16 / A) + 1 ]^(2)) - A = -28

A ( [(16)^2 / (A^2)] - (32 / A) ) = -28

(256 / A) - 32 = -28

A = 64

and since A + C = 100, C = 36

now go through the same method as shown above....

T(2)= 64e^(-k(2)) + 36 = 84

k = - ln[(84-36)/64) / 2 = ~0.1438

T(t) = 64e^(0.1438(t)) + 36

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• i'm assuming your M is the exterior temperature, so thus, right this is a thank you to unravel (a) one million) Distribute interior the ok dT/dt = kM - kT 2) flow the kT to the choice ingredient dT/dt + kT = kM 3) Use an integrating ingredient: intg is the fundamental µ = e^(intg(ok)dt) µ = e^kt 4) Multiply it in the time of and resolve for the remainder of the fundamental! e^kt*dT/dt + e^kt*kT = e^kt*kM combine: e^kt*T = intg(kM*e^kt)dt e^kt*T = M*e^kt + C 5) resolve for T T = M + C(e^-kt) M = exterior Temperature C = consistent ok = cost of substitute to unravel for (b), that is basically plug interior the values and resolve for the respecting ok and C. T = M + C(e^-kt) So via putting forward that the thermometer reads a hundred ranges initially, which potential T(0) = a hundred via after 6 minutes, temperature is eighty, so: T(6) = eighty The medium is likewise 70 ranges, indicating the preliminary temperature. Plug some values in and attempt it out. a hundred = 70 + C if t = 0, then e^-kt = one million subsequently C is 30. via then utilising the T(6) = eighty, you could resolve the ok. eighty = 70 + 30(e^-ok(6)) resolve for ok now. 10 = 30(e^-6k) one million/3 = e^-6k ln(one million/3) = -6k -ln(one million/3)/6 = ok subsequently, ok =~ 0.18 So the proper equation is: T = 70 + 30(e^-0.18t) Plug in 20 for t and resolve for T afterwards, T = 70.8º it extremely is logical because of the fact the temperature will ultimately attain the temperature of the median at 70º.

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• Anonymous

Write it as a differential equation and solve the equation. Hint: the result should be an exponential.

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• when you integrate the diff eq to find T(t) you will have 3 constants ,,k, C , and the integration constant. You are given 3 bits of data so using that find the constants.

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