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Exercise 5 · Q6

Q.A cake is removed from an oven at 250°F and left to cool at room temperature which is 70°F. After 30 minutes the temperature of the cake is 150°F. After how much time will it be 100°F?

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Newton's law gives T−70=180e−ktT-70=180e^{-kt}; the 30-minute cooling to 150°F fixes e−30k=49e^{-30k}=\tfrac{4}{9}, and then T=100T=100°F is reached at t≈66.3t\approx 66.3 minutes.

T(t)=Ts+(T0−Ts)e−ktT(t)=T_s+(T_0-T_s)e^{-kt} — Newton's law of cooling, where TsT_s = room temperature, T0T_0 = initial temperature, kk = cooling constant, tt = time in minutes.

  1. Set up. Ts=70∘T_s=70^\circF, T0=250∘T_0=250^\circF, so T−70=(250−70)e−kt=180e−ktT-70=(250-70)e^{-kt}=180e^{-kt}.
  2. Use the 30-minute data (T=150T=150°F at t=30t=30). 150−70=80=180e−30k⇒e−30k=80180=49150-70=80=180e^{-30k}\Rightarrow e^{-30k}=\dfrac{80}{180}=\dfrac{4}{9}.
  3. Find kk. −30k=log⁡ ⁣(49)=−0.81093⇒k=0.8109330=0.027031 min−1-30k=\log\!\left(\dfrac49\right)=-0.81093\Rightarrow k=\dfrac{0.81093}{30}=0.027031\ \text{min}^{-1}.
  4. Condition for 100°F. 100−70=30=180e−kt⇒e−kt=30180=16100-70=30=180e^{-kt}\Rightarrow e^{-kt}=\dfrac{30}{180}=\dfrac16. …

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