Question

A body cools from 80 °C to 50 °C in 5 minutes. Calculate the time it takes to cool from 60 °C to 30 °C. The temperature of the surroundings is 20 °C.

Answer 1

Here,
Initial temperature ( Ti) = 80°C
Final temperature ( Tf) = 50°C
Temperature of the surrounding ( To) = 20°C
t = 5 min
A/C to Newton's law of cooling

Rate of cooling ( dT/dt) = K[ (Ti+Tf)/2 - To]
( Tf - Ti)/t = K[ ( 80 + 50)/2 - 20]
( 80-50)/5 = K[ 65 - 20]
6 = K× 45
K = 6/45 = 2/15

in second condition,
initial temperature ( Ti) = 60°C
Final temperature ( Tf) = 30°C
Time taken for cooling is t
A/C Newton's law of cooling
( 60 - 30)/t = 2/15 [ (60+30)/2 -20]
30/t = 2/15 × 25
30/t = 50/15 = 10/3
t = 9 min

Answer 2

Explanation:

dT

​

=−K(T−T

s

​

)+T

i

​

n

∴ Here from newtons law of co0ling

dt

dT

​

=−k(

2

T

f

​

+T

i

​

​

−T

s

​

)

∴

△t

T

f

​

−T

i

​

​

=K[

2

T

f

​

+T

i

​

​

−T

s

​

]

∴K=

20

1

​

⇒=

10

60−T

​

=

20

1

​

[

2

60+T

​

−30]

=120−2T=

21

T+60

​

−30

2

5T

​

=120

T=48