. A cup of tea cools from 80 °C to 60 °C is one minute. The ambient temperature is 30 °C. In cooling from 60 °C to 50 °C. It will take :

From Newton's law of coolingdTdt=−KT−T0T1−T2t=KT1+T22−T080−6060=K80+602−3013=K×40    ….(1) and 60−50t=K60+502−3010t=K×25 …(2)From eqn (1) and (2)t30=4025⇒t=48sec