TS EAMCET 18-July-2022 Shift 1 Solved Paper

By Newton's law of cooling,
‌

−dT

dt

=b[T−T0]‌‌‌⋅⋅⋅⋅⋅⋅⋅(i)
wherè, ‌

−dT

dt

= Cooling rate,
T= Body's temperature,
T0= Surrounding temperature and b= Constant.
∴ From Eq. (i),

∫

i T

T‌

dT

T−T0

=−b‌

t

∫

0

dt
⇒‌‌T=T0+(Ti−T0)e−bt‌‌‌⋅⋅⋅⋅⋅⋅⋅(ii)
From the first condition of question,
T‌=40∘C
T0‌=10∘C
Ti‌=100∘C
t‌=10‌min
Putting these value in Eq. (ii),
40=10+(100−10)e−b×10
⇒‌‌‌

30

90

=e−10b
⇒‌‌‌

1

3

=e−10b
⇒‌‌ln(‌

1

3

)=−10b‌ln‌m
⇒‌‌ln‌1−ln‌3=−10b
⇒−ln‌3‌=−10b
⇒‌b‌=‌

ln‌3

10

=‌

1.1

10

Now, by next condition of question,
‌T=20∘C
‌Ti=70∘C
‌Ta=10∘C
From Eq. (ii),
‌20=10+(70−10)e−bt
‌‌

10

60

=e−‌

1.1

10

t
⇒‌‌ln(‌

1

6

)=−‌

11

6

t‌ln‌e
⇒‌‌ln‌l−ln‌6=−‌

1.1

10

t
⇒‌‌−ln‌6=‌

−11

100

t
⇒‌‌t=‌

100×ln‌6

11

=‌

100×1.8

11

=16.3‌min