TS EAMCET 03 May 2019 Shift 2 Solved Paper

From the question,
T1=70∘C,T2=40∘C,T0=20∘C‌ and ‌t=5min
The expression for Newton's law of cooling is given by,
‌

T1−T2

t

‌‌=k[‌

T1+T2

2

−T0]
‌

70−40

5

‌‌=k[‌

70+40

2

−20]
k‌‌=‌

6

35

........ (1)
If body cools from 60°C to 30C° in time t.
Then,
T1′=60∘C and T2′=30∘C
From Newton's law of cooling,
‌

T1−T2

t

‌‌=k[‌

T1+T2

2

−T0]
‌

60−30

t

‌‌=‌

6

35

[‌

60+30

2

−20]
‌

30

t

‌‌=‌

6

35

[25]
t‌‌=7min