Certain quantity of water cools from 70°C to 60°C in the first 5 minutes and to 54°C in the next 5 minutes. The temperature of the surroundings is
see full answer
Your Exam Success, Personally Taken Care Of
1:1 expert mentors customize learning to your strength and weaknesses – so you score higher in school , IIT JEE and NEET entrance exams.
An Intiative by Sri Chaitanya
a
45°C
b
20°C
c
42°C
d
10°C
answer is A.
(Unlock A.I Detailed Solution for FREE)
Best Courses for You
JEE
NEET
Foundation JEE
Foundation NEET
CBSE
Detailed Solution
Let Ts be the temperature of the surroundings. According to Newton's law of coolingT1−T2t=KT1+T22−Tsfor first 5 minutes,T1=70∘C,T2=60∘C,t=5 minutes ∴70−605=K70+602−Ts105=K65−TsT1=60∘C,T2=54∘C,t=5 minutes ∴ 60−545=K60+542−Ts65=K57−TsDivide eqn. (i) by eqn. (ii), we get53=65−Ts57−Ts285−5Ts=195−3Ts2Ts=90 or Ts=45∘C