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

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.