AP EAPCET 24-Aug-2021 Shift 1 Paper

Given that, for first condition,
Initial temperature, T1=52.5°C
Final temperature, Tv=47.5°C
Average temperature, Tav=

T1+T2

2

=

525+47.5

2

=50°C
For second condition,
Initial temperature, T1=47.5°C
Final temperature, T2=42.5°C
Average temperature, Tav=

T1+T2

2

=

47.5+42.5

2

=45°C
Let T0 be the temperature of surroundings.
Now, using expression of rate of cooling,
R=

ΔT

Δ

=−K(Tav−T0)
Substituting the values in above equation from 1‌st and 2nd conditions, we get

T1−T2

t1

=−K(Tav−T0)

52.5−47.5

5

=−K(50−T0) ...(i)
Similarly,

47,5−42.5

7.5

=−K(45−T0) ...(ii)
Dividing Eq. (i) by Eq. (ii), we get

52.5−47.5 5

47.5−42.5 7.5

=

−K(50−T0)

−K(45−T0)

⇒

5

5

×

7.5

5

=

50−T0

45−T0

⇒

3

2

=

50−T0

45−T0

T0=35°C
Hence, the temperature of surroundings is 35°C.