The rate at which a substance cools in moving air is proportional to the difference between the temperatures of the substance and that of the air. If the temperature of the air is 290 0 K and the substance cools from 370 0 K to 330 0 K in 10 minutes, when will the temperature be 295 0 K -
Text Solution
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Let T be the temperature of the substance at a time t then – 
∝ (T – 290) ⇒ 
= – k (T – 290)
Where k is constant of proportionality and negative sign denote rate of cooling.
or 
= – k dt
integrating, we get
= – k 
⇒ l n (T – 290) = – kt + l n c
⇒ 
= e –kt
or (T – 290) = ce –kt
If initially i.e., when t = 0 & T = 370
Then (370 – 290) = ce
∴ c = 80
∴ T – 290 = 80e k ... (1)
and for t = 10, T = 330
∴ 330 – 290 = 80e –10k ⇒ (40) = 80e –10k
⇒ 2 = e 10k
l n2 = 10 k ... (2)
To find t, when T = 295
From (1), 295 – 290 = 80e –kt
⇒ 
= e –kt ⇒ l n 16 = kt
or 4 l n 2 = kt ... (3)
Dividing (3) by (2) then 4 = 
∴ t = 40 minutes
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