Chapter 20, Problem 20.61AP

An aluminum rod 0.500 m in length and with a cross sectional area of 2.50 cm² is inserted into a thermally insulated vessel containing liquid helium at 4.20 K. The rod is initially at 3(H) K. (a) If one-halt 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, (b) 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 in liters per second after the lower half has reached 4.20 K? (Aluminum has thermal conductivity of 3 100 YV/m · K at 4.20 K; ignore its temperature variation. The density of liquid helium is 125 kg/m³.)

To determine

The volume of helium boil off by the time the inserted half cools is $4.20\text{K}$.

Answer to Problem 20.61AP

The volume of helium boil off by the time the inserted half cools to $4.20\text{K}$ is $17.2\text{L}$.

Explanation of Solution

Given info: The length of the aluminum rod is $0.500\text{m}$, the cross sectional area of the aluminum rod is $2.50{\text{cm}}^2$, the initial temperature of the aluminum rod is $300\text{K}$ and the final temperature of the aluminum rod and helium is $4.20\text{K}$.

Write the expression for the mass of the substance.

$m=\rho V$ (1)

Here,

$\rho$ is the density of the aluminum.

$V$ is the volume of the rod.

The expression for the volume of the rod is,

$V=Al$

Here,

$A$ is the area of the rod.

$l$ is the length of the rod.

Substitute $Al$ for $V$ in above equation.

$m=\rho Al$

For the mass of the aluminum rod:

Substitute $m_{\text{Al}}$ for $m$, $2700\text{kg}/{\text{m}}^3$ for $\rho$ and $0.500\text{m}$ for $l$ and $2.50{\text{cm}}^2$ for $A$ in the above equation.

$\begin{array}{c}{m}_{\text{Al}}=\left(2700\text{kg}/{\text{m}}^3\right)\left(2.50{\text{cm}}^2\times \frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\left(\frac{0.500\text{m}}{2}\right)\\ =0.16875\text{kg}\end{array}$

Thus, the mass of the aluminum rod is $0.16875\text{kg}$.

For the mass of the helium:

Substitute $m_{\text{He}}$ for $m$, and $125\text{kg}/{\text{m}}^3$ for $\rho$ in the equation (1).

$m_{\text{He}}=125\text{kg}/{\text{m}}^3\cdot V_{\text{He}}$

Here,

$V_{\text{He}}$ is the volume of the helium.

Write the expression for the amount of heat lost by the aluminum rod due to the change in temperature.

$Q_{\text{Al}}=m_{\text{Al}}{c}_{\text{Al}}\left(T_{\text{i}}-T_{\text{f}}\right)$

Here,

$m_{\text{Al}}$ is the mass of the aluminum rod.

$c_{\text{Al}}$ is the specific heat of the aluminum rod.

$T_{\text{f}}$ is the final temperature of the aluminum rod.

$T_{\text{i}}$ is the initial temperature of the aluminum rod.

Substitute $0.16875\text{kg}$ for $m_{\text{Al}}$, $900\text{J}/\text{kg}\cdot \text{K}$ for $c_{\text{Al}}$, $4.20\text{K}$ for $T_{\text{f}}$ and $300\text{K}$ for $T_{\text{i}}$ in above equation.

$\begin{array}{c}{Q}_{\text{Al}}=\left(0.16875\text{kg}\right)\left(900\text{J}/\text{kg}\cdot \text{K}\right)\left(\left(300\text{K}-4.20\text{K}\right)\right)\\ =44924.625\text{J}\end{array}$

Thus, the heat lost by the aluminum rod is $44924.625\text{J}$.

Write the expression for the heat gained by the helium.

$Q_{\text{He}}=m_{\text{He}}{L}_{\text{He}}$

Here,

$m_{\text{He}}$ is the mass of the helium.

$L_{\text{He}}$ is the latent heat of the helium.

Substitute $125\text{kg}/{\text{m}}^3\cdot V_{\text{He}}$ for $m_{\text{He}}$ and $2.09\times {10}^4\text{J}/\text{kg}$ for $L_{\text{He}}$ in the above equation.

$\begin{array}{c}{Q}_{\text{He}}=\left(125\text{kg}/{\text{m}}^3\cdot V_{\text{He}}\right)\left(2.09\times {10}^4\text{J}/\text{kg}\right)\\ =2612500\cdot V_{\text{He}}\end{array}$

From the law of the conservation of energy, the heat lost is equal to the heat gained.

Write the expression for the heat lost by the aluminum rod is equal to the heat gained by the helium.

$Q_{\text{He}}=Q_{\text{Al}}$

Substitute $2612.5\cdot V_{\text{He}}$ for $Q_{\text{He}}$ and $44924.625\text{J}$ for $Q_{\text{Al}}$ in the above equation.

$\begin{array}{c}2612500\cdot V_{\text{He}}=44924.625\text{J}\\ V_{\text{He}}\approx 0.0172{\text{m}}^3\times \frac{1000\text{L}}{1{\text{m}}^3}\\ =17.2\text{L}\end{array}$

Thus, the volume of helium boil off by the time the inserted half cools to $4.20\text{K}$ is $17.2\text{L}$.

Conclusion:

Therefore, the volume of helium boil off by the time the inserted half cools to $4.20\text{K}$ is $17.2\text{L}$.

To determine

The boil off rate of liquid helium when the lower half reached $4.20\text{K}$.

Answer to Problem 20.61AP

The boil off rate of liquid helium when the lower half reached $4.20\text{K}$ is $0.351\text{L}/\text{s}$.

Explanation of Solution

Given info: The length of the aluminum rod is $0.500\text{m}$, the cross sectional area of the aluminum rod is $2.50{\text{cm}}^2$, the initial temperature of the aluminum rod is $300\text{K}$ and the final temperature of the aluminum rod and helium is $4.20\text{K}$.

Write the expression for the rate at which energy is supplied to the rod.

$P=kA\left(\frac{T_{\text{i}}-T_{\text{f}}}{dx}\right)$

Here,

$k$ is the thermal conductivity of aluminum.

$A$ is the cross sectional area of the aluminum rod.

$dx$ is the change in length of the aluminum rod.

Substitute $3100\text{W}/\text{m}$ for $k$, $2.50{\text{cm}}^2$ for $A$, $\frac{0.500}{2}\text{m}$ for $dx$, $4.20\text{K}$ for $T_{\text{f}}$ and $300\text{K}$ for $T_{\text{i}}$ in above equation.

$\begin{array}{c}P=\left(3100\text{W}/\text{m}\right)\left(2.50{\text{cm}}^2\times \frac{{10}^{-4}{\text{m}}^2}{1{\text{cm}}^2}\right)\left(\frac{300\text{K}-4.20\text{K}}{\frac{0.500}{2}\text{m}}\right)\\ \approx 917\text{W}\end{array}$

Thus, the rate at which energy is supplied to the rod is $917\text{W}$.

Write the expression for the boil off rate of liquid helium after the lower half reaches $4.2\text{K}$.

$R=\frac{P}{\rho L_{\text{He}}}$

Substitute $917\text{W}$ for $P$, $125\text{kg}/{\text{m}}^3$ for $\rho$ and $2.09\times {10}^4\text{J}/\text{kg}$ for $L_{\text{He}}$ in above equation.

$\begin{array}{c}R=\frac{917\text{W}}{125\text{kg}/{\text{m}}^3\times 2.09\times {10}^4\text{J}/\text{kg}}\\ \approx 3.51\times {10}^{-4}{\text{m}}^3/\text{s}\times \frac{1000\text{L}}{1{\text{m}}^3}\\ =0.351\text{L}/\text{s}\end{array}$

Conclusion:

Therefore, the boil off rate of liquid helium when the lower half reached $4.20\text{K}$ is $0.351\text{L}/\text{s}$.