# A liquid cools from 55°C to 50°C in 5 minutes and it cools to 45°C in 7 minutes. Estimate the equation of temperature of surrounding.

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sciencesolve | Teacher | (Level 3) Educator Emeritus

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You should use the Netwon's law of temeparture change such that:

`(dT)/(dt) = k(T - E)`

`E ` represents the temperature of surrounding environment

`T(t)`  represents the temeprature of the liquid

`ln(T - E) = k*t + c => T - E = e^(k*t + c)`

At `t = 0` , the temeperature is of `55^o C`  such that:

`55^o - E = e^c*e^0 => 55^o - E = e^c`

At t = 5 min, the temperature is of `50^oC`  such that:

`50^o - E = e^c*e^(5k) => 50^o - E = (55^o - E)*e^(5k)`

`e^(5k) = (50^o - E)/(55^o - E) => e^k = root(5)((50^o - E)/(55^o - E))`

At `t =7 ` min, the temperature is of `45^oC`  such that:

`45^o - E = e^c*e^(7k)=> 45^o - E = (55^o - E)*((50^o - E)/(55^o - E))^(7/5) => (45^o - E)/(55^o - E) = ((50^o - E)/(55^o - E))^(7/5) `

`(45^o - E)/(55^o - E)- ((50^o - E)/(55^o - E))^(7/5) = 0`

Factoring out `1/(55^o - E)`  yields:

`1/(55^o - E)(45^o - E - (50^o - E)^(7/5)*(55^o - E)^(-2/5)) = 0`

`45^o - E - (50^o - E)^(7/5)*(55^o - E)^(-2/5) = 0`

`45^o - E = (50^o - E)^(7/5)*(55^o - E)^(-2/5)`

Hence, evaluating the relation that gives the temperature of surrounding environment yields `45^o - E = (50^o - E)^(7/5)*(55^o - E)^(-2/5).`

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