Question

A heat recovery system​ (HRS) is used to conserve heat from the surroundings and supply it to the Mars Rover. The HRS fluid loops use Freon as the working fluid. The instrumentation must be kept as a temperature greater than negative 65 degrees Fahrenheit​ [°F] to avoid damage. The temperature in the area of Mars where the rover is exploring is 192 kelvins​ [K]. If the system must remove 46.9 British Thermal Units​ [BTU] of​ energy, what volume of Freon is needed in units of liters​ [L]?

Answer 1

Answer: Volume of Freon = 12.23.$10^{3}$L

Explanation: First, the temperatures have to be in the same unit, so all the tmeperatures is transformed into Kelvin:

T₁ = 192K

T₂ = $\frac{5}{9}$. (F - 32)+273.15

T₂ = $\frac{5}{9}$. (- 65 - 32)+273.15

T₂ = 219.26K.

Second, BTU is an unit of energy. It can be transformed into Joules by the relation 1BTU = 1055.06J.

Q = 1055.06 . 46.9 = 49482.314J

The letter Q represents Heat and is calculated as Q = m.Cp.ΔT, where:

m is mass of the element;

Cp is the heat capacity specific for each element;

ΔT is the difference between the final and initial temperature;

Third, it will be needed the properties of the Freon:

Freon (CF2Cl2) = 120.91g/mol; Cp = 74J/mol.K; ρ = 1.49kg/m³

Fourth, we calculate the mass of Freon necessary for the remove of 46.9BTU of energy from the system:

Q = m.Cp.ΔT

m=$\frac{Q}{c(T-T_{0} )}$

m = $\frac{49482.314}{74(219.26-192)}$

m = 18228.214g or m=18.228Kg

Fifth, using the density of Freon, Volume can be found

ρ = $\frac{m}{V}$

V = m / ρ

V = 18.228 / 1.49

V = 12.23m³

As SI for density is Kg/m³, the volume found is in m³.

Cubic meters is related to Liters as the following: 1m³=1000l.

So, volume of Freon = 12.23.$10^{3}$L

The volume of Freon needed is 12.23.$10^{3}$L.