Question

A lake of water is at a temperature of 60∘F. The air temperature drops to 30∘F. Assume that Newton's law of cooling applies to the lake. If the water temperature drops 10∘F during the first day, how long will it take till the temperature of the lake is 40∘F.

Answer 1

Answer:

time = 2.7 days

Explanation:

Given:

Water temperature = 60°F

Air temperature drops = 30°F

1st day drop is = 10°F

Now according to Newtons law of cooling,

$\frac{dT}{dt}=-k.(T-T_{a})$

where dT is change in temperature

           dt is change in time

           k is coefficient of cooling

           T is temperature

           $T_{a}$ is ambient temperature = 30°F

∴$\frac{dT}{T-30}=-k.dt$

Integrating we get

ln(T-30) = -k.T+C -----------------------------(1)

Now when t = 0, T = 60°F

        when t = 1, T = 60-10

                              = 50°F

∴ln(60-30) = -k.0+C

  ln 30  = C

Now putting the value of C in (1)

ln (T-30) = -k.T+ln30

ln ( T -30) - ln 30 = -kt

ln ( t-30) / (30) = -k.T

Now at t - 1

ln (50-30) / 30 = -k x 1

ln ( 20/30 ) = -k

ln 2/3 = -k

∴$ln\left | \frac{(T-30)}{30}\right |= ln\left | \frac{2}{3} \right |\times t$

Let T = 40°F

$ln\left | \frac{(40-30)}{30}\right |= ln\left | \frac{2}{3} \right |\times t$

$ln\left | \frac{1}{3}\right |= ln\left | \frac{2}{3} \right |\times t$

$\frac{ln\left | \frac{1}{3} \right |}{ln\left | \frac{2}{3} \right |}= t$

t = $\frac{-1.0986}{-0.4054}$

 = 2.7 days

Answer 2

The lake of water by Newton's law of cooling will reduce due to temperature difference. The temperature of the lake of water reach to the 40 degree F in 2.7 (equal to 3) days.

What is Newton's law of cooling?

Newton's law of cooling states that the rate of heat transfer of a body is directly proportional to the temperature difference between the two bodies.

It can be given as,

$Q=-k(T-T_a)$

Here, (k) is the proportionality constant and $T_a$ is the ambient temperature.

Given information-

The initial temperature of the lake of water is 60 degree F.

The final temperature of the lake of water is 40 degree F.

The air temperature drops to 30 degree F.

The water temperature drops 10∘F during first day.

The above formula can be written as,

$\dfrac{dT}{T-30}=-kdt$

Integrate the above equation with respect to t as,

$\rm ln(T_i-30)=-kt+C\\ln(60-30)=-k\times0+C\\C=ln(30)$

Solve it again with the value of constant as,

$\rm ln(T_i-30)=-kt+ln(30)$

Let the above equation is 1

When the value of t is 1,

$\rm ln(T-30)=-k\times1+ln(30)\\k=-ln[ \dfrac{2}{3}]$

Put this value in the equation one as,

$\rm ln(T-30)=-[-ln(\dfrac{2}{3})]t+ln(30)\\ln[\dfrac{(T-30)}{30}]=ln(\dfrac{2}{3})]\times t$

The final temperature of the lake should be 40 degree F. Thus,

$\rm ln[\dfrac{(40-30)}{30}]=ln(\dfrac{2}{3})]\times t\\t=2.7$

Thus, the temperature of the lake of water reach to the 40 degree F in 2.7 (equal to 3) days.

Learn more about the Newton's law of cooling here;

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