Question 11.5: Boiling Liquid Helium Goal Apply the concept of latent heat ...
Boiling Liquid Helium
Goal Apply the concept of latent heat of vaporization to liquid helium.
Problem Liquid helium has a very low boiling point, 4.2 \mathrm{~K}, as well as a low latent heat of vaporization, 2.09 \times 10^{4} \mathrm{J} / \mathrm{kg}. If energy is transferred to a container of liquid helium at the boiling point from an immersed electric heater at a rate of 10.0 \mathrm{~W}, how long does it take to boil away 2.00 \mathrm{~kg} of the liquid?
Strategy Because L_{v}=2.09 \times 10^{4} \mathrm{~J} / \mathrm{kg}, boiling away each kilogram of liquid helium requires 2.09 \times 10^{4} \mathrm{~J} of energy. Joules divided by watts is time, so find the total energy needed and divide by the power to find the time.
Find the energy needed to vaporize 2.00 \mathrm{~kg} of liquid helium at its boiling point:
Q=m L_{v}=(2.00 \mathrm{~kg})\left(2.09 \times 10^{4} \mathrm{~J} / \mathrm{kg}\right)=4.18 \times 10^{4} \mathrm{~J}
Divide this result by the power to find the time:
\begin{aligned} & \Delta t=\frac{Q}{\mathscr{P}}=\frac{m L_{v}}{\mathscr{P}}=\frac{4.18 \times 10^{4} \mathrm{~J}}{10.0 \mathrm{~W}} \\ & \Delta t=4.18 \times 10^{3} \mathrm{~s}=69.7 \mathrm{~min} \end{aligned}
Remark Notice that no change of temperature was involved. During such processes, the transferred energy goes into changing the state of the substance involved.