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प्रश्न
If a body cools from 80°C to 50°C at room temperature of 25°C in 30 minutes, find the temperature of the body after 1 hour.
उत्तर
Let θ°C be the temperature of the body at time t minutes. Room temperature is given to be 25°C.
Then by Newton’s law of cooling, `(dθ)/dt` the rate of change of temperature, is proportional to (θ − 25).
i.e. `(dθ)/dt ∝ (θ - 25)`
∴ `(dθ)/dt` = −k(θ − 25), where k > 0
∴ `(dθ)/(θ - 25)` = −k dt
On integrating, we get
`int 1/(θ - 25)dθ = -k int dt + c`
∴ log (θ − 25) = −kt + c
Initially, i.e. when t = 0, θ = 80
∴ log (80 − 25) = −k × 0 + c ...(∴ c = log 55)
∴ log (θ − 25) = −kt + log 55
∴ log (θ − 25) − log 55 = −kt
`log ((θ - 25)/55) = -kt` ...(1)
Now, when t = 30, θ = 50
∴ `log ((50 - 25)/55) = - 30"k"`
∴ k = `- 1/30 log (5/11)`
∴ (1) becomes, `log ((θ - 25)/55) = t/30 log (5/11)`
When t = 1 hour = 60 minutes, then
`log ((θ - 25)/55) = 60/30 log (5/11)`
`log ((θ - 25)/55) = 2 log (5/11)`
∴ `log ((θ - 25)/55) = log (5/11)^2`
∴ `(θ - 25)/55 = (5/11)^2`
∴ `(θ - 25)/55 = 25/121`
∴ `θ - 25 = 55 xx 25/121`
∴ `θ - 25 = 125/11`
∴ `θ = 125/11 + 25`
∴ `θ = (125 + 275)/11`
∴ `θ = 400/11`
∴ θ = 36.36
∴ θ the temperature of the body will be 36.36°C after 1 hour.
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Then the rate of increase of p is `"dp"/"dt"` which is proportional to p.
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`int "dp"/"p" = "k"int "dt"`
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∴ log p – log 100000 = kt
∴ `log ("P"/100000)` = kt ......(i)
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∴ `log (200000/100000)` = 25k
∴ k = `square`
∴ equation (i) becomes, `log("p"/100000) = square`
When p = 400000, then find t.
∴ `log(400000/100000) = "t"/25 log 2`
∴ `log 4 = "t"/25 log 2`
∴ t = `25 (log 4)/(log 2)`
∴ t = `square` years
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∴ `("dp")/"dt" prop "p"`
∴ Differential equation can be written as `("dp")/"dt" = "kp"`
where k is constant of proportionality.
∴ `("dp")/"p" = "k.dt"`
On integrating we get
`square` = kt + c ...(i)
(i) Where t = 0, p = 1,00,000
∴ from (i)
log 1,00,000 = k(0) + c
∴ c = `square`
∴ log `("p"/(1,00,000)) = "kt"` ...(ii)
(ii) When t = 25, p = 2,00,000
as population doubles in 25 years
∴ from (ii) log2 = 25k
∴ k = `square`
∴ log`("p"/(1,00,000)) = (1/25log2).t`
(iii) ∴ when p = 4,00,000
`log ((4,00,000)/(1,00,000)) = (1/25log2).t`
∴ `log 4 = (1/25 log2).t`
∴ t = `square ` years
