TS EAMCET 4 May 2018 Shift 2 Solved Paper

According to the Newton’s law of cooling,
mC(

T1−T2

t

)=k(

T1+T2

2

−T0)
For first case,
Substitute the values.
mC(

100−70

300

)=k(

100+70

2

−30)
mC[

30

300

]=k[

170

2

−30]
mC[

1

10

]=k[85−30]
mC[

1

10

]=k(55) …… (I)
For second case,
Substitute the values.
mC(

100−80

t"

)=k(

100+80

2

−30)
mC[

20

t"

]=k[90−30]
mC[

20

t"

]=k[60] …… (II)
On solving equations (I) and (II) we get,
t"=3.05‌min