CGPET 2019 Solved Paper

Given, T1=60°C,T2=50°C
T0=25°,t2=t1=10min
According to Newton's law of cooling,

T1−T2

t

=k(

T1+T2

2

−T0)
⇒

60−50

10

=k(

60+50

2

−25)
⇒ 1=k(55−25)⇒k=

1

30

.......(i)
Suppose if temperature of body θ in next 10min, then

50−θ

10

=k(

50+θ

2

−25) .........(ii)
From Eqs. (i) and (ii)

50−θ

10

=

1

30

(

50+θ

2

−25)
⇒

50−θ

10

=

1

30

(

50+θ−50

2

)

50−θ

10

=

1

30

×

θ

2

⇒ 50−θ=

θ

6

⇒ 300−6θ=θ
⇒ 7θ=300⇒θ=

300

7

=42.8°C