Heat and Thermal ExpansionHard
Question
A cup of tea cools from 80oC to 60oC in one minute.the ambient temperature is 30oC. In cooling from 60oC to 50oC. will take :-
Options
A.50 s
B.90 s
C.60 s
D.48 s
Solution
Newton′s law of Cooling 
= k[θ - θ0]
θ0 = surrounding′s temperature
⇒
.....(1)
and
.....(2)
⇒ t = 48 sec
= k[θ - θ0]θ0 = surrounding′s temperature
⇒
.....(1)and
.....(2)⇒ t = 48 sec
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