A body initially at a all 100 degree centigarde cools to 60 degree centigarde in 5 minutes and to 40 degree centigarde in 10 minutes . What is the temperature of surrounding? What will be the temperature in 15 minutes?
Answers
Answer:
According newton's law of cooling,
T=T
s
+(T
o
−T
s
)e
−kt
where, T= temp. at any time t
T
o
=initial temp.
T
s
=surrounding temp.
t=time
k=cooling constant
since body cools initially from 60℃ to 50℃ in 10 min.
so from above equation,
50=20+(60-20)e
−k(10)
⇒
4
3
=e
−10k
................eqution(1)
we need to find the temp. at the end of next 10 min. i.e. 20 min. from start
so again using newton's law of cooling formula,
T=T
s
+(T
o
−T
s
)e
−kt
⇒T=20+(60-20)e
−k(20)
⇒T=20+(40){e
−10k
}
2
..................equation (2)
from equation (1) and (2), we get
⇒T=20+(40)(
4
3
)
2
⇒T=42.5 ℃
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