Question

Two large containers A and B of the same size are filled with different fluids. The fluids in containers A and B are maintained at 0° C and 100° C, respectively. A small metal bar, whose initial temperature is 100° C, is lowered into container A. After 1 minute the temperature of the bar is 90° C. After 2 minutes (since being lowered into container A) the bar is removed and instantly transferred into the other container. After 1 minute in container B the temperature of the bar rises 10°. How long, measured from the start of the entire process, will it take the bar to reach 99.4° C? (Round your answer to two decimal places. Assume the final temperature being asked for is reached while the bar is container B.)

t = ___ min

Answer 1

The DE for Newton's Law of cooling /warming is dT/dt=k(T -$T_{m})$ --> dT/$T- T_{m}$ =kdt

$\therefore \int \frac{dT}{T-T_{m}}=\int kdt$

=$ln\left | T-T_{m} \right |=kt+c_{1}$

$T-T_{m}=e^{kt+c_{1}}$ =T=$T_{m}+e^{c_{1}}e^{kt}$ =$T_{m}+Ce^{kt}$

For the first process:

Since the surrounding medium is container A->$T_{m}=0^{0}C$

Now applying the initial condition T(0)= $100^{0}C$ we get 100=0+C=>C=100=> T= $100e^{kt}$

since the temperature decreased to $90^{0}$ C after one minute,

90=100$e^{k1}$ => $e^{k1}$ =90/100 => k1=ln(9/10)=-0.10536=>T=100$e^{-0.10536t}$

thus after 2 minutes we have T=100$e^{(-0.10536*2)}$ =$81^{0}C$

For hte second process:

Since the surrounding medium is container B=> $T_{m}= 100^{0}C$ .

now we have initial temperature =$81^{0}C$

thus applying the initial condition we get 81=100+C => C= 81-100 =>C=-19.

therefore T=100-19$e^{kt}$

since the temperature increased by $10^{0}$ C after one minute

81+10=100-19$e^{k2}$

$19e^{k2}=9 =>e^{k2}=9/19 =>k2=ln(9/19)=-0.7472$

Hence T=100-$19e^{-0.7472t}$

thus for the bar to reach 99.4 we have

99.4=100-$19e^{-0.7472t} =>e^{-0.7472t}=0.6/19$

therefore -0.7472 t=ln (0.6/19)

Hence t= 4.62 minutes

thus the total time from the entire process to reach 99.4 is t= t1+t2=2+4.62=6.62 minutes.