Key Idea From Newton's law of cooling, the time taken (t) by a body to cool from T1 to T2 when placed in a medium of temperature T0 can be calculated from relation tT1−T2=K1(2T1+T2−T0) When the object cool from 80∘C to 70∘C in 12 minutes, then from Newton's law of cooling, 1280−70= K1(280+70−25)[∵T0=25∘C]65= K150 Similarly, when object cool from 70∘C to 60∘C we get t70−60= K1(270+60−25)t10= K140 Divide Eq. (i) and (ii), we get 65×10t=4050⇒t=45×12=15 minutes
An object kept in a large room having air temperature of 25° C takes 12 minutes to cool from 80° C to 70° C. The time taken to cool for the same…
NEET UG 2019 — Physics Thermodynamics
An object kept in a large room having air temperature of 25∘C takes 12 minutes to cool from 80∘C to 70∘C. The time taken to cool for the same object from 70∘C to 60∘C would be nearly
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
10 min
- B
12 min
- C
20 min
- D
15 min
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