An aluminum rod 0.500 m in length and with a cross-sectional area of 2.50 cm^2 is inserted into a thermally insulated vessel containing liquid helium at 4.20 K (B.P. for helium). The rod is initially at 300 K. (๐_๐ด๐=900 ๐ฝ/๐๐ยฐ๐ถ, ๐ฟ_(๐ฃ,๐ป๐)=2.09ร10^4 ๐ฝ/๐๐)
If one half of the rod is inserted into the helium, how many liters of helium boil off by the time the inserted half cools to 4.20 K? Assume the upper half does not yet cool.
If the circular surface of the upper end of the rod is maintained at 300 K, what is the approximate boil-off rate of liquid helium after the lower half has reached 4.20 K? (Aluminum has thermal conductivity of 3100 W/m ยท K at 4.20 K; ignore its temperature variation. The densities of liquid helium and aluminum are 0.125 gใ ๐๐ใ^(โ3) and 2.7 gใ ๐๐ใ^(โ3) respectively.)
To find the number of liters of helium boiled off, we first need to calculate the heat transferred from the aluminum rod to the helium.
The heat transferred can be calculated using the equation:
Q = mcฮT,
where Q is the heat transferred, m is the mass of the aluminum rod, c is the specific heat capacity of aluminum, and ฮT is the change in temperature.
The mass of the aluminum rod can be calculated using the equation:
m = ฯV,
where ฯ is the density of aluminum and V is the volume of the aluminum rod.
The volume of the aluminum rod is given by:
V = A * L,
where A is the cross-sectional area of the aluminum rod and L is its length.
Substituting the values given, we have:
V = (2.50 cm^2) * (0.500 m) = 0.0125 m^3.
The mass of the aluminum rod is then:
m = (2.7 g/cm^3) * (0.0125 m^3) = 0.03375 kg.
Now we can calculate the heat transferred:
Q = (0.03375 kg) * (900 J/(kgยทยฐC)) * (300 K - 4.20 K) = 8,402.625 J.
The heat transferred is equal to the heat absorbed by the liquid helium. The heat absorbed by the liquid helium can be calculated using the equation:
Q = mLv,
where Q is the heat absorbed, m is the mass of the liquid helium boiled off, and Lv is the latent heat of vaporization of helium.
The mass of the liquid helium boiled off is then:
m = Q / Lv = 8,402.625 J / (2.09ร10^4 J/kg) = 0.401422 kg.
The number of liters of helium boiled off can be calculated using the equation:
V_H = m / ฯ_H,
where V_H is the volume of the liquid helium boiled off and ฯ_H is the density of liquid helium.
Substituting the values given, we have:
V_H = (0.401422 kg) / (0.125 g/cm^3) = 3211.376 cm^3.
Converting cm^3 to liters using the conversion factor 1 cm^3 = 0.001 liters, we get:
V_H = 3211.376 cm^3 * (0.001 liters/cm^3) = 3.211376 liters.
Therefore, approximately 3.211 liters of helium boil off by the time the inserted half of the rod cools to 4.20 K.
Now let's calculate the boil-off rate of liquid helium after the lower half of the rod has reached 4.20 K.
The rate of heat transfer from the aluminum rod to the helium can be calculated using the equation:
Q = kAฮT / ฮดx,
where Q is the rate of heat transfer, k is the thermal conductivity of aluminum, A is the cross-sectional area of the aluminum rod, ฮT is the temperature difference between the aluminum rod and the helium, and ฮดx is the thickness of the aluminum rod.
The thickness of the aluminum rod is half of its length:
ฮดx = (0.500 m) / 2 = 0.250 m.
Substituting the values given, we have:
Q = (3100 W/(mยทK)) * (2.50 cm^2) * (300 K - 4.20 K) / (0.250 m) = 2,737,500 W.
The rate of heat transfer is equal to the rate of heat absorbed by the liquid helium:
Q = mLv,
where Q is the rate of heat absorbed, m is the rate of mass flow of liquid helium, and Lv is the latent heat of vaporization of helium.
The rate of mass flow of liquid helium can be calculated using the equation:
m = Q / Lv = 2,737,500 W / (2.09ร10^4 J/kg) = 130.994 kg/s.
The rate of mass flow can be converted to the rate of volume flow using the equation:
V_H = m / ฯ_H,
where V_H is the rate of volume flow of liquid helium and ฯ_H is the density of liquid helium.
Substituting the values given, we have:
V_H = (130.994 kg/s) / (0.125 g/cm^3) = 10,479.952 cm^3/s.
Converting cm^3/s to liters/s using the conversion factor 1 cm^3/s = 0.001 liters/s, we get:
V_H = 10,479.952 cm^3/s * (0.001 liters/cm^3) = 10.479952 liters/s.
Therefore, the approximate boil-off rate of liquid helium is approximately 10.48 liters/s after the lower half of the rod has reached 4.20 K.
To solve this problem, we need to consider the heat transfer from the aluminum rod to the liquid helium. We can use the formula for heat transfer:
Q = mcฮT
Where Q is the heat transfer, m is the mass of the aluminum rod, c is the specific heat capacity of aluminum, and ฮT is the change in temperature.
1. Calculate the mass of the aluminum rod:
Given: Length of the rod = 0.500 m
Cross-sectional area = 2.50 cm^2 = 2.50 * 10^-4 m^2
So, the volume of the rod is: Volume = Length * Cross-sectional area
Volume = 0.500 m * 2.50 * 10^-4 m^2 = 1.25 * 10^-4 m^3
The density of aluminum is 2.7 g/cm^3 = 2.7 * 10^3 kg/m^3
Mass of the aluminum rod is: Mass = Volume * Density
Mass = 1.25 * 10^-4 m^3 * 2.7 * 10^3 kg/m^3 = 0.3375 kg
2. Calculate the heat transfer from the aluminum rod to the helium:
Given: Initial temperature of the rod, T_initial = 300 K
Final temperature of the rod, T_final = 4.20 K
Specific heat capacity of aluminum, c_Al = 900 J/kgยฐC
Change in temperature, ฮT = T_final - T_initial = 4.20 K - 300 K = -295.80 K
The heat transfer is: Q = mcฮT
Q = 0.3375 kg * 900 J/kgยฐC * (-295.80 K) = -90,426.75 J
Since the rod is inserted halfway, only half of the rod's length is in contact with the helium. Therefore, only half of the heat transferred will cause the helium to boil off.
3. Calculate the liters of helium that boil off:
Given: Latent heat of vaporization of helium, L_v,He = 2.09 * 10^4 J/kg
The mass of helium that boils off is: m_He = Q / L_v,He
m_He = -90,426.75 J / 2.09 * 10^4 J/kg = -4.33 kg
The density of helium is 0.125 g/cm^3 = 0.125 * 10^3 kg/m^3
Volume of helium that boils off is: V_He = m_He / density_He
V_He = -4.33 kg / 0.125 * 10^3 kg/m^3 = -34.64 * 10^-3 m^3
Since 1 L = 10^-3 m^3, the volume of helium that boils off is approximately 34.64 L.
Therefore, approximately 34.64 liters of helium boil off by the time the lower half of the rod cools to 4.20 K.
4. Calculate the approximate boil-off rate of liquid helium:
Given: Thermal conductivity of aluminum, k_Al = 3100 W/m ยท K
The rate of heat transfer is given by: Q_dot = k_Al * A * (T_2 - T_1) / L
Where Q_dot is the heat transfer rate, A is the cross-sectional area of the rod, T_1 is the temperature on the surface, T_2 is the temperature at the other end, and L is the length of the rod.
We are given that T_1 = 300 K, T_2 = 4.20 K, and A = 2.50 cm^2 = 2.50 * 10^-4 m^2.
The length L = 0.500 m, but only half of this length is cooled.
So, L = 0.500 m / 2 = 0.250 m
The heat transfer rate is: Q_dot = 3100 W/m ยท K * 2.50 * 10^-4 m^2 * (4.20 K - 300 K) / 0.250 m
Q_dot = 3100 W/m ยท K * 2.50 * 10^-4 m^2 * (-295.80 K) / 0.250 m
Q_dot = -368,868 W
Since the heat transfer rate is negative, this means heat is flowing from the helium to the aluminum rod. Therefore, the approximate boil-off rate of liquid helium is approximately 368,868 W.
Please note that the negative sign indicates that heat is being transferred from the helium to the rod. The magnitude of the heat transfer rate indicates the rate at which the helium is boiling off.