a body cools from 62 degree Celsius to 50 degree Celsius in 10 minutes and 42 degree Celsius in next 10 minutes then the temperature of surrounding is
Answers
Answer:
By Newton's law of cooling
θ
2
−θ
1
dθ
2
=kdt
where k is inversely proportional to product of mass and specific heat of the water, θ
1
and θ
2
are surrounding and average water temperatures and dt is the time interval.
Given that for dθ
2
=62
o
C−50
o
C=12
o
C, θ
1
=26
o
C, θ
2
=
2
62+50
=56
o
C, dt=10min
And for the first ten minutes
θ
2
−θ
1
dθ
2
=kdt
56−26
12
=k×10
k=
300
12
min
o
C
For the next ten minutes
θ
2
−θ
1
dθ
2
=kdt
2
θ
2
+θ
f
−26
θ
2
−θ
f
=
300
12
×10
2
50+θ
f
−26
50−θ
f
=
300
12
×10
50−θ
f
=0.2(θ
f
−2)
θ
f
=41.3
o
C
Explanation:
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Answer:
According to Newton's law of cooling
θ2−θ1dθ2=kdt
where k is inversely proportional to product of mass and specific heat of the water, θ1 and θ2 are surrounding and average water temperatures and dt is the time interval.
Given that for dθ2=62oC−50oC=12oC, θ1=26oC, θ2=262+50=56oC, dt=10min
And for the first ten minutes
θ2−θ1dθ2=kdt56−2612=k×10k=30012minoC
For the next ten minutes