# 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???

### 6 Answers

- BernardLv 41 decade agoFavorite Answer
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.

Edit: Sorry, Misread the question

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

and C = 36˚ Centigrade

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- ?Lv 43 years ago
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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- Anonymous1 decade ago
Write it as a differential equation and solve the equation. Hint: the result should be an exponential.

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- ted sLv 71 decade ago
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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- Anonymous1 decade ago
this is your homework bud do it by yourself

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