Heat Transfer Calculation in a Thermally Insulated Vessel with Liquid Helium

a. How many liters of helium boil off when half of the aluminum rod is inserted into the liquid helium at 4.20 K?

a. 0.0818 L

b. What is the approximate boil-off rate of liquid helium in liters per second after the lower half of the rod has reached 4.20 K?

b. 0.073 L/s

Answer:

The number of liters of helium that boil off when half of the rod is inserted can be calculated using the heat transfer formula and the density of helium. The boil-off rate of liquid helium can be determined using the heat transfer formula and the time taken for the rod to cool.

Explanation: To determine the number of liters of helium that boil off when half of the rod is inserted into the helium, we need to calculate the heat transfer between the rod and the helium. We can use the formula Q = mcΔT, where Q is the heat transfer, m is the mass of the helium, c is the specific heat capacity of helium, and ΔT is the change in temperature.

The heat transfer will cause the helium to boil off, so we can convert the heat transfer into liters of helium using the density of helium. In this case, the number of liters of helium that boil off is approximately 0.0818 L when half of the aluminum rod is inserted into the liquid helium at 4.20 K.

To calculate the boil-off rate when the lower half of the rod has reached 4.20 K, we can use the formula Q/t, where Q is the heat transfer and t is the time taken. The heat transfer can be calculated as before, and the time taken can be determined by finding the time it takes for the rod to cool from 3(H) K to 4.20 K at a given rate of heat transfer. The approximate boil-off rate of liquid helium in this scenario is 0.073 L/s.