Question

dit The rate at which a body cools also depends on its exposed surface area S. If Sis a constant, then a modification of (2) given in Section 3.1, dr kS(T-T). where k < and Te is a constant. Suppose that two cups A and Bare filled with coffee at the same time. Initially, the temperature of the coffee is 66* C. The exposed surface area of the coffee in cup is twice the surface area of the coffee in cup A After 30 min the temperature of the coffee in cup A is 38°C. If T-22°C, then what is the temperature of the coffee in cup after 30 min? (Round your answer to two decimal places.) 27.50

Answer 1

boely codes solni Geven rati at which ат KS (T-Tm) Where Tm = 22°c. at where let SA and SB are the exposed surface according to question SB = 2.SA 2 By can 6 at KS at T-Tm F = ksat T-22 ou entegrating ln (1-22) = KS ttc But initially t = 66°c Egiven) en (66-22) = KS XO + C (= ln (44) 80 en (1-22) = K St + ln (44) In (1-22) - lu694) = KSE lu (1-22 44 Kst Now for Cup A, T= 38c & S = SA t for Cup B T= & S= SB ? 60 lu(12) = kset ☺ ln (38-22 ) = KSAt 4 Scanned with CamScanner

Now Diui ③ 44, we get dividing ln (1-22) 44 SB al SA en Go 16 44 ln (1-22) lu/T-22 44 2 SA = ; using (2 en/16 SA 44 ln (1-22 =2 en lau) T-22 2 こ cent 16 44 44 T-22 = 256 5.8181 44 =) T= 22 + 5,8181 T=27.8181 °C CSScanned with CamScanner