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Exercise 10.8 · Q8

Q.At 10.00 A.M. a woman took a cup of hot instant coffee from her microwave oven and placed it on a nearby kitchen counter to cool. At this instant the temperature of the coffee was 180∘F180^\circ F, and 1010 minutes later it was 160∘F160^\circ F. Assume that the constant temperature of the kitchen was 70∘F70^\circ F.

(i) What was the temperature of the coffee at 10.15 A.M.? [log⁡911=−0.6061]\left[\log\dfrac{9}{11}=-0.6061\right]
(ii) The woman likes to drink coffee when its temperature is between 130∘F130^\circ F and 140∘F140^\circ F. Between what times should she have drunk the coffee? [log⁡611=−0.2006]\left[\log\dfrac{6}{11}=-0.2006\right]
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Newton's law of cooling with Tm=70T_m=70; fix CC at t=0t=0 and kk from the t=10t=10 reading, then evaluate at t=15t=15 for part (i) and invert for T=130,140T=130,140 in part (ii).

Step 1. Set up. T−70=Ce−ktT-70=Ce^{-kt} (writing the decay explicitly with k>0k>0). At t=0, T=180t=0,\ T=180: C=110C=110.

Step 2. Fix kk from T(10)=160T(10)=160. 160−70=90=110e−10k ⟹ e−10k=911 ⟹ k=−110ln⁡911=110ln⁡119≈0.02007160-70=90=110e^{-10k}\ \Longrightarrow\ e^{-10k}=\dfrac{9}{11}\ \Longrightarrow\ k=-\dfrac1{10}\ln\dfrac9{11}=\dfrac1{10}\ln\dfrac{11}9\approx0.02007.

Step 3. (i) Temperature at t=15t=15 (10:15 A.M.). T(15)−70=110e−15k=110(e−10k)1.5=110(911)1.5≈110(0.7402)≈81.4T(15)-70=110e^{-15k}=110\left(e^{-10k}\right)^{1.5}=110\left(\dfrac9{11}\right)^{1.5}\approx110(0.7402)\approx81.4. So T(15)≈151.4∘FT(15)\approx151.4^\circ F. …

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