A thermometer is taken from an inside room to the outside, where the air tempera

Question

A thermometer is taken from an inside room to the outside, where the air temperature is 10 degree F. After 1 minute the thermometer reads 60 degree F, and after 5 minutes it reads 35 degree F. What is the initial temperature of the inside room? () degree F A tank contains 180 liters of fluid in which 20 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 3 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t. A(t) = g

Explanation / Answer

Newton's Law of Heat Transfer: rate of change of temperature (heat flow) is proportional to difference between temperature of object and temperature of surroundings (ambient)

EDIT: I used T0 for ambient temperature but that was confusing even for me, so I altered it to Ta
T is temperature
t is time
k is constant of proportionality (thermal resistance, analogous to a capacitor draining or charging through a resistor)
dT(t)/dt = -k(T(t) - Ta)
Ta is ambient, 10°F
T(1) = 60 °F
T(5) = 35 °F
We want to find T(0), the initial temperature
This is a classical exponential equation with solution
T(t) = Ta + (T(0) - Ta) exp(-k t)
60 = 10 + (T(0) - 10) exp(-k)
35 = 10 + (T(0) - 10) exp(-5 k)

50 = (T(0) - 10) exp(-k) --------------------- (1)
25 = (T(0) - 10) exp(-5 k) --------------------- (2)
Divide (1) by (2)
2 = exp(4 k)
k = (ln 2)/4
Resubstitute into (1) and solve
50 = (T(0) - 10) exp(-(ln(2)/4)
T(0) = 50 (2^(1/4)) + 10
Initial temperature = 69.5 °F to one decimal place --- ANSWER

2) Fluid is being pumped in and pumped out at same rate

So there is always 100L of fluid in tank
At time t, the amount of salt in tank is A(t) (in grams)
So concentrations of salt at time t is A/100 g/L

Amount of brine pumped in: 3L (with salt concentration = 1 g/L)
Amount of salt pumped in: 3L * 1g/L = 3g

Amount of brine pumped out at time t: 3L (with salt concentration = A/100 g/L)
Amount of salt pumped out at time t: 3L * A/100 g/L = 3A/100 g

dA/dt = 3 - 3A/100
100 dA/dt = 300 - 3A
100/(300 - 3A) dA = dt
(-100/3) ln|3A-300| = t + C
ln|3A-300| = -3t/100 + C .... where
3A - 300 = C e^(-3t/100) ..... where C = e^C
3A = 300 + C e^(-3t/100)

A = 100 + (1/3) C e^(-3t/100)

Initially, brine contains 20 g of salt
A(0) = 20
100 + (1/3)C e^0 = 20
C = -240

A(t) = 300 - 240e^(-3t/100) .

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