Question

a cup of tea cools from 80-79.9 in 5 minutes.if temperature of surrounding is 20 degree celsius then how much time it will take to cool from70-69.9degree celsius assume Newton law of cooling is valid here​

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

$-\frac{dT}{dt} = k(T - T_c)$

now we have

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

here we know that

$-\frac{dT}{dt}$ = Rate of decrease in temperature

$T_s$ = surrounding temperature

T = temperature of tea

now integrating both sides we have

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

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

now we know that

$T_2 = 79.9$

$T_1 = 80$

$T_s = 20$

so we have

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

$k = 3.34 \times 10^{-4}$

now again from above equation we have

$T_2 = 69.9$

$T_1 = 70$

so we have

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

$t = 6 min$

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Topic : Newton's law of cooling

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