According to Stefan's law of radiation, the absolute temp. T of a body cooling in a medium at constant absolute temp. (T sub m) or Tm is given by:
dT/dt=k(T^4-T^4sub m) Solve the differential equation
This is a variables separable first order differential equation.
dT/dt = k(T^4 - Tm^4) ----> INT dT/(T^4 - Tm^4) = INT k dt
Integrate each side separately. You will need to split the left side into partial fractions.
A/(T - Tm) + B/(T + Tm) + CT/(T^2 + Tm^2) + D/(T^2 + Tm^2)
Can you finish from there? Remember that Tm is a constant. The first three terms on the left will integrate to natural logarithms and the fourth to arctan.
So then I completed it....is this correct?
Could you tell me if this is right? I got A ln(t)+Bln(t)+Cln(t)/2+1/T^2m[tan^-1 (T/Tm)]=kt+C....
then I divided both sides by t (basically just put all that over t)....is this my final answer k=this answer?
dT/dt=k(T^4-T^4sub m) Solve the differential equation
This is a variables separable first order differential equation.
dT/dt = k(T^4 - Tm^4) ----> INT dT/(T^4 - Tm^4) = INT k dt
Integrate each side separately. You will need to split the left side into partial fractions.
A/(T - Tm) + B/(T + Tm) + CT/(T^2 + Tm^2) + D/(T^2 + Tm^2)
Can you finish from there? Remember that Tm is a constant. The first three terms on the left will integrate to natural logarithms and the fourth to arctan.
So then I completed it....is this correct?
Could you tell me if this is right? I got A ln(t)+Bln(t)+Cln(t)/2+1/T^2m[tan^-1 (T/Tm)]=kt+C....
then I divided both sides by t (basically just put all that over t)....is this my final answer k=this answer?
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Starting from dT/(T^4 - (T_m)^4) = k dt:
Let a = T_m for simplicity.
By partial fractions,
1/(T^4 - a^4) = A/(T - a) + B/(T + a) + (CT + D)/(T^2 + a^2).
Clearing denominators:
1 = A(T + a)(T^2 + a^2) + B(T - a)(T^2 + a^2) + (CT + D)(T^2 - a^2)
Letting T = a yields 1 = (2a^3)A ==> A = 1/(2a^3)
Letting T = -a yields 1 = (-2a^3)B ==> B = -1/(2a^3)
Letting T = ai yields 1 = (Cai + D)(-2a^2) ==> C = 0 and D = -1/(2a^2).
Hence, (1/(2a^3)) [1/(T - a) - 1/(T + a) - a/(T^2 + a^2)] = k dt
==> 1/(T - a) - 1/(T + a) - a/(T^2 + a^2) = 2ka^3 dt.
Integrate both sides:
ln(T - a) - ln(T + a) - arctan(T/a) = 2ka^3 t + C.
==> ln [(T - a)/(T + a)] - arctan(T/a) = 2ka^3 t + C.
Assuming that T(0) = T₀ (initial temperature), we find that
ln(T₀ - a)/(T₀ + a)] - arctan(T₀/a) = C.
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For your second part:
Assume that T - T_m is negligible compared to T_m itself.
Working with the original DE,
dT/dt = k(T^4 - (T_m)^4)
.........= k[((T - T_m) + T_m)^4 - (T_m)^4], expanding in powers of T - T_m
.........= k[(T - T_m)^4 + 4T_m (T - T_m)^3 + 6(T_m)^2 (T - T_m)^2 + 4(T_m)^3 (T - T_m)]
Any power of T - T_M is especially negligible, so ignoring these terms yields
dT/dt = k[0 + 4(T_m)^3 (T - T_m)]
........= (4k(T_m)^3) * (T - T_m),
which is essentially Newton's Law of cooling with constant 4k(T_m)^3.
I hope this helps!
Let a = T_m for simplicity.
By partial fractions,
1/(T^4 - a^4) = A/(T - a) + B/(T + a) + (CT + D)/(T^2 + a^2).
Clearing denominators:
1 = A(T + a)(T^2 + a^2) + B(T - a)(T^2 + a^2) + (CT + D)(T^2 - a^2)
Letting T = a yields 1 = (2a^3)A ==> A = 1/(2a^3)
Letting T = -a yields 1 = (-2a^3)B ==> B = -1/(2a^3)
Letting T = ai yields 1 = (Cai + D)(-2a^2) ==> C = 0 and D = -1/(2a^2).
Hence, (1/(2a^3)) [1/(T - a) - 1/(T + a) - a/(T^2 + a^2)] = k dt
==> 1/(T - a) - 1/(T + a) - a/(T^2 + a^2) = 2ka^3 dt.
Integrate both sides:
ln(T - a) - ln(T + a) - arctan(T/a) = 2ka^3 t + C.
==> ln [(T - a)/(T + a)] - arctan(T/a) = 2ka^3 t + C.
Assuming that T(0) = T₀ (initial temperature), we find that
ln(T₀ - a)/(T₀ + a)] - arctan(T₀/a) = C.
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For your second part:
Assume that T - T_m is negligible compared to T_m itself.
Working with the original DE,
dT/dt = k(T^4 - (T_m)^4)
.........= k[((T - T_m) + T_m)^4 - (T_m)^4], expanding in powers of T - T_m
.........= k[(T - T_m)^4 + 4T_m (T - T_m)^3 + 6(T_m)^2 (T - T_m)^2 + 4(T_m)^3 (T - T_m)]
Any power of T - T_M is especially negligible, so ignoring these terms yields
dT/dt = k[0 + 4(T_m)^3 (T - T_m)]
........= (4k(T_m)^3) * (T - T_m),
which is essentially Newton's Law of cooling with constant 4k(T_m)^3.
I hope this helps!
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