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A cup of tea cools from 80°c to 60°c in 60 seconds. The ambient temperature is 30°c. In cooling from 60°c to 50°c, it will take?

A cup of tea cools from 80°c to 60°c in 60 seconds. The ambient temperature is 30°c. In cooling from 60°c to 50°c, it will take?

Grade:12

2 Answers

Mukti Patel
11 Points
5 years ago
I calculate and my answer is 60s. It has formula of substance temperature decrease T1 to T2 and duration for this is time t the formula is:-{ T2-T1/t = - k(T1 +T2/2 - Ts)
Khimraj
3007 Points
5 years ago
According to newton’s law of cooling
log(T2 – T)/(T1 – T) = Kt
log(80 – 30)/(60 – 30) = K*60
log5/3 = 60K
Then K = (1/60)log5/3
and final condition is
log(60 -30)/(50-30) = Kt
log3/2 = [(1/60)log5/3]t
So t = 60log(3/2)/log(5/3).
Hope it clears.

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