Question

Liquid helium and liquid nitrogen are both used as coolants; He(l) boils at 4.21 K, and ${\mathbf{N}}_{\mathbf{2}}\left(\mathbf{l}\right)$ boils at 77.35 K. The specific heat of liquid helium near its boiling point is $\mathbf{4}\mathbf{.}\mathbf{25}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$, and the specific heat of liquid nitrogen near itsboiling point is $\mathbf{1}\mathbf{.}\mathbf{95}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{g}}^{\mathbf{-}\mathbf{1}}$. The enthalpy of vaporization of He(l) is $\mathbf{25}\mathbf{.}\mathbf{1}{\mathbf{Jg}}^{\mathbf{-}\mathbf{1}}$, and the enthalpy of vaporization of ${\mathbf{N}}_{\mathbf{2}}\left(\mathbf{l}\right)$is$\mathbf{200}\mathbf{.}\mathbf{3}{\mathbf{Jg}}^{\mathbf{-}\mathbf{1}}$ (these data are calculated from the values in Appendix F). Discuss which liquid is the better coolant (on a per-gram basis) nearits boiling point and which is better atits boiling point.

Short answer

The heat required to melt 1.00 g of ice is 314 J.

Step-by-step solution

The given Information.

Liquid helium boils at 4.21 K.

Liquid nitrogen boils at 77.35 K.

The specific heat of liquid helium near its boiling point is$4.25{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$.

The specific heat of liquid nitrogen near its boiling point is$1.95{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$.

The enthalpy of vaporization of He(l) is $25.1{\mathrm{Jg}}^{-1}$.

The enthalpy of vaporization of ${\mathrm{N}}_2\left(\mathrm{l}\right)$ is $200.3{\mathrm{Jg}}^{-1}$.

The coolant.

The temperature of a particular system is decreased or simply controlled using the material that is termed coolant. The characteristics of a good coolant include high thermal capacity, low cost, and chemically inert.

The liquid that is a better coolant.

The liquid that is a better coolant near its boiling point;

As the specific heat of liquid helium near its boiling point is $4.25{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$and the specific heat of liquid nitrogen near its boiling point is $1.95{\mathrm{JK}}^{-1}{\mathrm{g}}^{-1}$.

The specific heats are provided based on a per-gram basis.

It is observed that the specific heat of liquid helium is greater than liquid nitrogen near the boiling point.

Hence, 1 g of liquid helium will remove more heat as its specific heat is greater than liquid nitrogen, and the liquid helium is a better coolant near its boiling point.

Also, as the boiling point, enthalpy of vaporization is a governing factor.

If the enthalpy of vaporization is large, a large amount of heat is extracted, and the liquid performs better as a coolant on a per-gram basis.

The enthalpy of vaporization of He is$25.1{\mathrm{Jg}}^{-1}$ and the enthalpy of vaporization of$N_2$ is $200.3{\mathrm{Jg}}^{-1}$.

Hence, 1 g of liquid nitrogen will remove heat as its enthalpy of vaporization is larger than liquid helium, and the liquid nitrogen is a better coolant (on a per-gram basis) at its boiling point.