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Worked Examples · Example 15

Q.A cake is taken out from an oven when its temperature has reached 185°F and is placed on a table in a room whose temperature is 75°F. If the temperature of the cake reaches 150°F after half an hour, what will be its temperature after 45 minutes?

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Newton cooling T=75+110e−ktT=75+110e^{-kt}: the 30-min datum gives e−30k=75110e^{-30k}=\tfrac{75}{110}, so at t=45t=45, T≈136.9∘T\approx136.9^\circF.

dTdt=−k (T−Ts)  ⇒  T=Ts+(T0−Ts)e−kt\dfrac{dT}{dt}=-k\,(T-T_s)\;\Rightarrow\;T=T_s+(T_0-T_s)e^{-kt}, where TsT_s = surrounding temp, T0T_0 = initial temp, k>0k>0 cooling constant, tt = time.

Given: Ts=75∘T_s=75^\circF, T0=185∘T_0=185^\circF, T(30 min)=150∘T(30\text{ min})=150^\circF.

  1. Model: T=75+(185−75)e−kt=75+110 e−ktT=75+(185-75)e^{-kt}=75+110\,e^{-kt}.
  2. Use T=150T=150 at t=30t=30: 150=75+110e−30k  ⇒  110e−30k=75  ⇒  e−30k=75110=0.6818150=75+110e^{-30k}\;\Rightarrow\;110e^{-30k}=75\;\Rightarrow\;e^{-30k}=\dfrac{75}{110}=0.6818.
  3. Required temperature at t=45t=45: T=75+110 e−45kT=75+110\,e^{-45k}. …

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