Anonymous
Anonymous asked in Science & MathematicsMathematics · 1 decade ago

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

Relevance
  • 1 decade ago
    Favorite 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

    • Login to reply the answers
  • ?
    Lv 4
    3 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.

    • Login to reply the answers
  • Anonymous
    1 decade ago

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

    • Login to reply the answers
  • ted s
    Lv 7
    1 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.

    • Login to reply the answers
  • How do you think about the answers? You can sign in to vote the answer.
  • 1 decade ago

    Go and do your homework yourself...

    • Login to reply the answers
  • Anonymous
    1 decade ago

    this is your homework bud do it by yourself

    • Login to reply the answers
Still have questions? Get your answers by asking now.