Question

A boiling 100 c solution is set on a table
Where the room temperature is assumed to
be constant at 20 c the solution cooled
to 60 c after 5 minutes
(i) find a formula for the temperature (T)
of the solution , t minutes after it is
placed on table.
(1) Determine how long it will takes for
the solution to cool to 22°C.

Answer 1

Answer explanation:

First of all u can't assume anything. The question is wrong. U simply just make life easier by assuming things. MArk brainliest.Iknow u won't so thanks for the points

Answer 2

Answer:

i) $T(t)=T_s\pm (T_o-T_s)e^{-0.138(t)}$

ii) Time taken to reach $22^oC$ is 26.7 minutes.

Explanation:

i) Using Newton's law of cooling:

$\frac{-dT}{dT}\propto (T_o-T_s)$

where, $T_o$ is the temperature of the object.

$T_s$ is the temperature of the object.

$\implies T(t)=T_s\pm (T_o-T_s)e^{-kt}$----(1)

here, $k$ is proportionality constant

Given: After time t = 5min; T(5) = $60^oC$;

Temperature of solution: $T_o=100^oC$;

Temperature of surrounding:  $T_s=20^oC$

Putting in values in (1)

$\implies 60=20+(100-20)e^{-k(5)}$

$\implies 40=80+e^{-5k}$

$\implies 0.5=e^{-5k}$

$\implies \frac{ln(2)}{5}=k$

$\implies k=0.138$

the formula for the temperature (T) of the solution after t minutes is,

$\implies T(t)=T_s\pm (T_o-T_s)e^{-0.138(t)}$

ii) Now, After time t T(t) = $22^oC$

Putting in values in (1)

$\implies 22=20+(100-20)e^{-0.138(t)}$

$\implies 2=80*e^{-0.138(t)}$

$\implies \frac{1}{40}=e^{-0.138(t)}$

$\implies \frac{ln(40)}{0.38}=t$

$\implies t=26.7$

Time taken to reach $22^oC$ is 26.7 minutes.

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