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A Body is Heated at 110°C and Placed in Air at 10°C. After 1 Hour Its Temperature is 60°C. How Much Additional Time is Required for It to Cool to 35°C? - Mathematics and Statistics

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Sum

A body is heated at 110°C and placed in air at 10°C. After 1 hour its temperature is 60°C. How much additional time is required for it to cool to 35°C?

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Solution

Let `theta^circ "C"` he the temperature of the body at any time t.
Temperature of air is 10o C i.e `theta_circ = 10`
According to Newton's law of cooling, we have

`(d theta)/dt prop theta-theta_@`

`therefore (d theta)/dt =-k( theta-theta_@),k>0`

`therefore ("d"theta)/"dt" = - "K" (theta - 10)`

`therefore 1/(theta-10) "d"theta =-kdt`

Integrating both sides, we get


`int 1/(theta-10) "d"theta=- "k" int "dt"`


`therefore "log" (theta - 10) = -"Kt" + "C"`


When t = 0 , θ = 110° C

∴ log (110 - 10) = - K × 0 + C        ∴ C = log 100

∴ log (θ - 10) = - Kt + log 100


`"log" ((theta - 10)/100) = - "Kt"`


Also θ = 60° C ; when t = 1


`therefore "log"  ((60-10)/100) = - "K" xx 1`


`therefore "K" = - "log" (1/2)`


`therefore "log" ((theta - 10)/100) = "t"  "log" (1/2)`


when θ = 35° C , then 


`"log" ((35-10)/100) = "t"  "log" (1/2)`


`"log" (25/100) = "t log" (1/2)`


`"log" (1/4) = "t log" (1/2)`


- log 4 = - t log 2


t = `("log" 4)/("log" 2) = 2`


The additional time required for body to cool to 35° C = (2 - 1) = 1 hour

Concept: Differential Equations - Applications of Differential Equation
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