Question

Answer me this question.
​

Answer 1

Answer: Average temperature

=94+862=1802=90∘C

Excess temperature =90∘−20∘=70∘C

Rate of fall in temperature,

dθ1dt=−[94−862]=−[82]

=−4∘Cmin−1

dθdt∝(θ−θ0)=−k(θ−θ0)

−4=k×70…(i)

In second case,

Average temperature

=74+662=1402=70∘C

Excess temperature

=70∘−20∘C=50∘C

Rate of fall in the temperaturre

dθ2dt=−[74−66t2]

=−8∘Ct2min−1

∴−8t2=k×50…(ii)

On dividing Eq. (i) by Eq. (ii), we get

t22=75

t2=145=2.8min