Question

18. of tea cools from 80°C to 79.9°C in 5 minute. If temperature of surroundings is 20°C. Then how much time will it take to cool from 70°C to 69.9°C? Assume Newton's law of cooling is valid here (1) 5 minute (2) 6 minute (3) 7 minute (4) 8 minute th Tower, 8, Pusa Road, New Delhi-110005 Ph.011-47623456​

Answer 1

Answer:

It will take 6 minutes to cool from 70 degree celsius to 69.9 degree celsius

Explanation:

As per Newton's law of cooling we know that

-\frac{dT}{dt} = k(T - T_c)−

dt

dT

=k(T−T

c

)

now we have

\frac{dT}{T - T_c} = k dt

T−T

c

dT

=kdt

here we know that

-\frac{dT}{dt}−

dt

dT

= Rate of decrease in temperature

T_sT

s

= surrounding temperature

T = temperature of tea

now integrating both sides we have

-ln(\frac{T_2 - T_s}{T_1 - T_s}) = kt−ln(

T

1

−T

s

T

2

−T

s

)=kt

T_2 = T_s + (T_1 - T_s)e^{-kt}T

2

=T

s

+(T

1

−T

s

)e

−kt

now we know that

T_2 = 79.9T

2

=79.9

T_1 = 80T

1

=80

T_s = 20T

s

=20

so we have

79.9 = 20 + (80 - 20)e^{-5k}79.9=20+(80−20)e

−5k

k = 3.34 \times 10^{-4}k=3.34×10

−4

now again from above equation we have

T_2 = 69.9T

2

=69.9

T_1 = 70T

1

=70

so we have

69.9 = 20 + (70 - 20)e^{-kt}69.9=20+(70−20)e

−kt

t = 6 mint=6min