Questions
A body initially at 80°C cools to 64°C in 5 minutes and to 52°C in 10 minutes. The temperature of the body after 15 minutes will be
detailed solution
Correct option is A
According to Newton's law of cooling, θ1−θ2t=Kθ1+θ22−θ0 For first process: (80−64)5=K80+642−θ0 …(i)For second process: (80−52)10=K80+522−θ0 …(ii)For third process: (80−θ)15=K80+θ2−θ0 …(iii)On solving Eqs. (i) and (ii), we getK=115 and θ0=24∘CPutting these values in Eq. (iii), we getθ=42.7∘C.Talk to our academic expert!
Similar Questions
A body cools down from 60°C to 55°C in 30 s. Using Newton’s law of cooling, calculate the approximate time taken by same body to cool down from 5 5°C
to 50°C. Assume that the temperature of surroundings is 45°C.
799 666 8865
support@infinitylearn.com
6th Floor, NCC Building, Durgamma Cheruvu Road, Vittal Rao Nagar, HITEC City, Hyderabad, Telangana 500081.
free study material
JEE Mock Tests
JEE study guide
JEE Revision Notes
JEE Important Questions
JEE Sample Papers
JEE Previous Year's Papers
NEET previous year’s papers
NEET important questions
NEET sample papers
NEET revision notes
NEET study guide
NEET mock tests
