5. An object has an initial temperature of 10°C. At t = 0 it is placed into a container whose temperature oscillates according to A(t) = 5 cos 2rt, where t is in hours. Apply Newton's law of heat conduction (T'(t) = 2(A(t) – T(t))) to determine the temperature of the object as a function of time. Identify the transient and steady-state portions of your solution. Use Wolfram Alpha if you can't remember how to do the integral! On the exam I'll give one with an easier integral. Answ: T(t) = (cos 2nt + T sin 2nt) + (10 – z)e-2t, Ttrans = (10 – )e-", Tss = (cos 2nt + T sin 2nt)) 1+7 %3D 5. %3D

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5. An object has an initial temperature of 10°C. At t = 0 it is placed into a container whose
temperature oscillates according to A(t) = 5 cos 2nt, where t is in hours. Apply Newton's law of
heat conduction (T'(t) = 2(A(t) – T(t))) to determine the temperature of the object as a function
of time. Identify the transient and steady-state portions of your solution. Use Wolfram Alpha if you
can't remember how to do the integral! On the exam I'll give one with an easier integral.
Answ: T(t) = (cos 2nt + T sin 2nt) + (10 – )e-2t,
Ttrans = (10 – )e-24,
Tss = (cos 2at + n sin 27t))
1+7
Transcribed Image Text:5. An object has an initial temperature of 10°C. At t = 0 it is placed into a container whose temperature oscillates according to A(t) = 5 cos 2nt, where t is in hours. Apply Newton's law of heat conduction (T'(t) = 2(A(t) – T(t))) to determine the temperature of the object as a function of time. Identify the transient and steady-state portions of your solution. Use Wolfram Alpha if you can't remember how to do the integral! On the exam I'll give one with an easier integral. Answ: T(t) = (cos 2nt + T sin 2nt) + (10 – )e-2t, Ttrans = (10 – )e-24, Tss = (cos 2at + n sin 27t)) 1+7
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