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Exercise 10.8 · Q7

Q.Water at temperature 100∘C100^\circ C cools in 1010 minutes to 80∘C80^\circ C in a room temperature of 25∘C25^\circ C. Find

(i) The temperature of water after 2020 minutes
(ii) The time when the temperature is 40∘C40^\circ C [log⁡e1115=−0.3101; log⁡e5=1.6094]\left[\log_e\dfrac{11}{15}=-0.3101;\ \log_e5=1.6094\right]
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Newton's law of cooling with Tm=25T_m=25; fix CC from t=0t=0 and kk from t=10t=10, then answer both parts using that same kk.

Step 1. Set up. T−Tm=CektT-T_m=Ce^{kt} with Tm=25T_m=25. At t=0, T=100t=0,\ T=100: C=75C=75.

Step 2. Fix kk from T(10)=80T(10)=80. 80−25=75e10k ⟹ e10k=5575=111580-25=75e^{10k}\ \Longrightarrow\ e^{10k}=\dfrac{55}{75}=\dfrac{11}{15}, so k=110ln⁡1115k=\dfrac1{10}\ln\dfrac{11}{15} (negative, as expected for cooling). Using the given ln⁡1115=−0.3101\ln\dfrac{11}{15}=-0.3101: k=−0.03101k=-0.03101.

Step 3. (i) Temperature after 20 minutes. T(20)−25=75e20k=75(e10k)2=75(1115)2=75×121225≈40.3T(20)-25=75e^{20k}=75\left(e^{10k}\right)^2=75\left(\dfrac{11}{15}\right)^2=75\times\dfrac{121}{225}\approx40.3. So T(20)≈25+40.3=65.3∘CT(20)\approx25+40.3=65.3^\circ C. …

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