Question

An apparatus that liquefies helium is in a room maintained at 300 K. If the helium in the apparatus is at 4.0 K, what is the minimum ratio${\mathbf{Q}}_{\mathbf{to}}\mathbf{/}{\mathbf{Q}}_{\mathbf{from}}$, where${\mathbf{Q}}_{\mathbf{to}}$is the energy delivered as heat to the room and${\mathbf{Q}}_{\mathbf{from}}$is the energy removed as heat from the helium?

Short answer

The ratio of the energy delivered as heat to the room to that of the energy removed as heat from the helium is 75

Step-by-step solution

The given data

${\mathrm{Q}}_{\mathrm{to}}$is the energy delivered as heat to the room

${\mathrm{Q}}_{\mathrm{from}}$is the energy removed as heat from helium

Room temperature$T_{to}=300K$

The temperature of the system$T_{from}=4K$

Understanding the concept of entropy change to get energy

By using the concept of change in entropy for a complete cycle is zero. i.e. entropy is a state function, by using the equation, we can find the ratio${\mathbf{Q}}_{\mathbf{t}\mathbf{o}}\mathbf{/}{\mathbf{Q}}_{\mathbf{f}\mathbf{r}\mathbf{o}\mathbf{m}}$.

Formula:

The entropy change using the second law of thermodynamics,

$∆\mathrm{S}=\frac{\mathrm{Q}}{\mathrm{T}}$ (1)

Calculation of the ratio Qto/Qfrom

The net entropy change of the system due to energy transfer by the body and to the body can be given using equation (1) as:

$$\begin{gathered} ∆\mathrm{S}=∆{\mathrm{S}}_{\mathrm{to}}+∆{\mathrm{S}}_{\mathrm{from}} \\ =\frac{{\mathrm{Q}}_{\mathrm{to}}}{{\mathrm{T}}_{\mathrm{to}}}-\frac{{\mathrm{Q}}_{\mathrm{from}}}{{\mathrm{T}}_{\mathrm{from}}}...........................\left(2\right) \end{gathered}$$

$∆{\mathrm{S}}_{\mathrm{from}}$is negative because heat is removed from the substance.

As we know, entropy is a state function total change in entropy must be zero i.e.$∆\mathrm{S}=0$

Thus, the equation of entropy change gives the relation of energy and temperature as follows:

$\frac{{\mathrm{Q}}_{\mathrm{to}}}{{\mathrm{T}}_{\mathrm{to}}}=\frac{{\mathrm{Q}}_{\mathrm{from}}}{{\mathrm{T}}_{\mathrm{from}}}$

(where${\mathrm{Q}}_{\mathrm{to}}$ is the heat delivered to the room,${\mathrm{Q}}_{\mathrm{from}}$ is heat removed from the helium,${\mathrm{T}}_{\mathrm{to}}$ is the room temperature, and${\mathrm{T}}_{\mathrm{from}}$ is the helium temperature.)

$$\begin{gathered} \frac{{\mathrm{Q}}_{\mathrm{to}}}{{\mathrm{Q}}_{\mathrm{from}}}=\frac{300\mathrm{K}}{4\mathrm{K}} \\ =75.0 \end{gathered}$$

Hence, the required value of the ratio is 75.0