Question
The temperature of a body falls from 40∘C to 36∘C in 5 minutes when placed in a surrounding of constant temperature 16∘C . Find the time taken for the temperature of the body to become 32∘C.
Answer: Option B
:
B
As the temperature differences are small, we can use Newton's law of cooling.
dθdt=−k(θ−θ0)dθθ−θ0=−kdt ...(i)
Where k is a constant,θ is the temperature of the body at time t and θ0 = 16∘C is the temperature of the surrounding. We have,
∫360c400cdθθ−θ0=−k(5min)In360c−160c400c−160c=−k(5min)
or,k=−In(5/6)5min
or,If t be the time required for the temperature to fall from 360C to 320C then by (i),
∫320c360cdθθ−θ0=−ktIn320c−16c360c−160=−In(5/6)t5mint=In(4/5)In(5/6)×5min=6.1min
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:
B
As the temperature differences are small, we can use Newton's law of cooling.
dθdt=−k(θ−θ0)dθθ−θ0=−kdt ...(i)
Where k is a constant,θ is the temperature of the body at time t and θ0 = 16∘C is the temperature of the surrounding. We have,
∫360c400cdθθ−θ0=−k(5min)In360c−160c400c−160c=−k(5min)
or,k=−In(5/6)5min
or,If t be the time required for the temperature to fall from 360C to 320C then by (i),
∫320c360cdθθ−θ0=−ktIn320c−16c360c−160=−In(5/6)t5mint=In(4/5)In(5/6)×5min=6.1min
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