TS EAMCET 04 May 2019 Shift 1 Solved Paper

According to the expression of Newton's law of cooling,
−‌

dT

dt

=T0+(T1−T0)e−kt
Substitute ‌

T(t)−T0

T1−T0

for e−kt in the above equation
−‌

dT

dt

‌‌=T0+(T1−T0)e−kt
t‌‌=‌

1

k

‌ln‌

T1−T0

T(t)−T0

For T1=0∘C
t=‌

1

k

‌ln‌

−T0

T(t)−T0

Substitute mcT(t) for Q(t) in the above equation.
t=‌

1

k

‌ln‌

−Q0

Q(t)−Q0

According to the condition given in the question the t1 and t2 is calculated as,
t1=‌

1

k

‌ln(‌

−Q0

60 100 Q0−Q0

)
And also,
t2=‌

1

k

‌ln(‌

−Q0

80 100 Q0−Q0

)
The ratio ‌

t2

t1

is calculated as
‌

t2

t

=‌

ln( −Q0 80Q0−100Q0 100 )

ln(‌ ‌ −Q0 60Q0−100Q0 100 )

=‌

ln( −Q0×100 −20Q0 )

ln(‌ −Q0×5 −2Q0 )