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