A certain substance cools from 370 K to 330 K in 10 minutes, when the temperature of
air is 290 K. The temperature after 40 minutes will be
(a)270 K
(b)325 K
(c)300 K
(d)295 K
please explain the answer
Answers
Answer:
According to Newton’s law of cooling, the rate at which a substance cools in moving air is proportional to the difference between the temperature of the substance and that of the air. If the temperature of the air is 290°C and the substance cools from 370°C to 330°C in 10 minutes, find when the temperature will be 295°C.
Newton's law of cooling is Q/t∝(T _ s−T _ f)Q/t∝(T
s
−T
f
)
Solving for it, we get;
T(t)=T_ S+(T_ 0–T_ S)e –ktT(t)=T
S
+(T
0
–T
S
)e–kt
where; t is the time ;
T_o=T
o
= initial temperature of substance
T_s=T
s
= temperature of surrounding fluid
In this case
T _ s=290°CT
s
=290°C ;T _ o=370°C;T
o
=370°C
For t=10 minutes and T=330°C---(given)
To find: t for T=295°CT=295°CTofind:tforT=295°CT=295°C
330=290+80e^{-10k}330=290+80e −10k330=290+80e
−10k
330=290+80e−10k
\implies k =ln2/10⟹k=ln2/10⟹k=ln2/10⟹k=ln2/10
295=290+80e −kt295=290+80e−kt
⟹e ^{kt} =16⟹t=4ln2/k⟹e
kt
=16⟹t=4ln2/k
⟹t=40min. (Answer)⟹t=40min.(Answer)