Question

QUESTION 6 The temperature of a cup of coffee when it is poured is 81 °C. After 3 minutes, in a room that is at 20°C, the temperature of the coffee drops to 69 °C. find the temperature after 5 minutes? write your answer to the nearest Two decimal places.

Answer 1

By Newton's law of cooling, -K(T- Ta Ta) dT at Here, Ta= tooom temperature -20°c ) 1. © becomes, dT at -K(T-20) dT 3-K at T- 20 पा dT T- 20 s- kat d(T-20) =-kt 0 + ln e T-20 [in C = integral const. → Intf-20) = -Kt+Inc 10(5-20) 2 in 1 T-20 :) -kt TT T-20 ekt -kt T= 20+ cé 2 Acording to the problem, when t-o then T=81'0

.: 81 = 20 + e e-K.O → 81-20 -Cl [:eº=1] * e=61 2 becomes, - Kt T= 20+ ble 3 Now abtoh 3 minutes temperature. 06 the cotee is 69°c. . t=3 then. T= 69 ii (3 ③ -3K 69= 20 + 61 ē -BK & blé 49 e 3k 49 -3K e 61 -3K = ln 49 61 -3K - en l 49 GI K = GI a K = ln ( 4 ) - (49) tof en (49) π t .. 32 T-20 +Gle t=5 then. .: After 5 minutes, T= 20.461.e 61.2 50 (49) .5 = 62.3422 atter 5 minutes .: Temperature is | 62.34°C (Answer)