Question

Newton's Law of Cooling states that the rate of change of the temperature of a cooling body is proportional to the difference between the temperature of the body and the constant temperature of the medium surrounding the body, Apply this law to the following problem: At 10:00 am, 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°F, and 10 minutes later it was 160°F. If we assume that the constant temperature of the kitchen was 70°F, what was the temperature of the coffee at 10:15 am? F F 10(7+ PVT) ° (10+29) 10 (vi ° 7(10+?!) 7 10 11+ 27 F F None of them

Answer 1

Option A is correct

t = 5 min >10:15 T = 2 t-to t-10min Giveni 10:00 >IDAD To = 180° F T'=160 F Surrounding temp. Ts = 70°f Acc. to Newton's Law of cooling & I- Ts = (To-Ts) ekt 160-70 180- 70 In 10.8182) = -K (10). Ine K = 0.0201 min -K(10) ox

for t=5min, (T' -Is) ekt T-Ts F. I-70 160-70 -0.0201 X5 e T = 151.41°F Now checking options, 10177 22.0) 10/7 + 8.14) - 151.41 F. Hence, option (4) is correct.