Transmission Of Heat Question 176
A cup of tea cools from $ 80^{o}C $ to $ 60^{o}C $ in one minute. The ambient temperature is $ 30^{o}C $ . In cooling from $ 60^{o}C $ to $ 50^{o}C $ it will take
[MP PMT 1995; UPSEAT 2000; MH CET 2002]
Options:
A) $ 30\ \sec onds $
B) $ 60\ \sec onds $
C) $ 90\ \sec onds $
D) $ 50\ \sec onds $
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Answer:
Correct Answer: D
Solution:
$ \frac{80-60}{1}=K( \frac{80+60}{2}-30 ) $
Therefore $ K=\frac{1}{2} $ Again $ \frac{60-50}{t}=\frac{1}{2}( \frac{60+50}{2}-30 ) $
Therefore $ t=0.8\times 60=48 $ sec.
BETA
