Method-1 Using exact law of cooling ‌T−Ts=(T0−Ts)e−Kt ‌‌ Case-I: ‌(40−10)=(60−10)e−7K ‌30=50e−7K ‌‌ Case-II: ‌(T−10)=(40−10)e−7K‌ or ‌T−10=30e−7K ‌‌ Dividing ‌(2)‌ by ‌(1) ‌‌
T−10
30
=‌
30
50
‌⇒T−10=‌
30×30
50
=18 ‌‌ or ‌T=28∘C Methode-2 Using newton's average law of cooling ‌‌