Q.

A cup of tea cools from 80oC to 60oC in 1 min. The ambient temperature is 30oC. In cooling from 60oC to 50oC it will take

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a

30 s

b

60 s

c

90 s

d

48 s

answer is D.

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Detailed Solution

According to Newton's law of cooling   θ1−θ2t∝θ1+θ22−θFor the first condition80−6060∝80+602−30  …..(i)and for the second condition60−50t∝60+502−30 ……(ii)By solving Eqs. (i) and (ii), we get t = 48 s.
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