Chapter 13, Problem 82GP

(a)

To determine

To Find: The total mass of air.

Answer to Problem 82GP

The total mass of air inside the house at 20 ⁰C is $925.76\text{ kg}$ .

Explanation of Solution

Given:

Volume of house $V=770{\text{ m}}^{\text{3}}$ .

The density of air ${\rho }_{air}=1.29{\text{ kg / m}}^{\text{3}}at0\text{ °C}$ .

Temperature $T=20\text{ °C }$ .

Coefficient of thermal expansion of air ${\beta }_{air}=3.4\times {10}^{-3}{\text{ °C}}^{\text{-1}}$ .

Formula used:

The total mass of air inside the house is the difference between the mass of air at 0 ⁰C and mass of air at 20 ⁰C. The following formula will be helpful t find the total mass of air inside the house.

Mass of air inside the house $\text{= mass at }0\text{ °C - mass at }20\text{ °C}$ .

Density $\rho =\frac{\text{Mass }(m)}{\text{Volume }(V)}$ .

Mass $(m)=\text{ Density }(\rho )\times \text{Volume }(V)$ .

  $\Delta V_{20\text{°C}}=V\times \beta \times \Delta T$

.Calculation:

Substitute the given values,

mass at $\text{0}°\text{c}$ = $1.29\times 770=993.3\text{ kg}$

  $\Delta V=770\times 3.4\times {10}^{-3}\times 20=52.36{\text{ m}}^{\text{3}}$

Mass at $\text{20}°\text{c}$ = $1.29\times 52.36=67.54\text{ kg}$

The total mass of air inside the house = $993.3-67.54=925.76\text{ kg}$

Conclusion:

Thus, the total mass of air inside the house at ${20}^{\circ }C$ is $925.76\text{ kg}$ .

(b)

To determine

To Find: The total mass of air when the temperature decreases.

Answer to Problem 82GP

The total mass of air enters the house and the mass of air when the temperature decreases to -10 ⁰C are $101.32\text{ kg}$ .

Explanation of Solution

Given:

  $\begin{array}{l}\text{Volume of house }V=770{\text{ m}}^{\text{3}}\\ \text{Density of air }{\rho }_{air}=1.29{\text{ kg / m}}^{\text{3}}at0\text{ °C}\\ \text{Temperature }T=20\text{ °C }-(-10\text{ °C})=30\text{ °C}\\ \text{Coefficient of thermal expansion of air }{\beta }_{air}=3.4\times {10}^{-3}{\text{ °C}}^{\text{-1}}\end{array}$

Formula used:

Find the total mass of air when temperature decreases to -10 ⁰C by the following way,

  $\begin{array}{l}\text{Density }\rho =\frac{\text{Mass }(m)}{\text{Volume }(V)}\\ \text{Mass }(m)=\text{ Density }(\rho )\times \text{Volume }(V)\\ \Delta V_{20\text{°C}}=V\times \beta \times \Delta T\end{array}$

Calculation:

Substitute the given values,

  $\begin{array}{l}\Delta V=770\times 3.4\times {10}^{-3}\times 30=78.54{\text{ m}}^{\text{3}}\\ \text{Mass at }-10\text{ °C }=1.29\times 78.54=101.32\text{ kg}\end{array}$

At -10 ⁰C, the mass of air is $101.32\text{ kg}$ , which is more dense compare to 20 ⁰C and so that the mass of air enters the house.

Conclusion:

Thus, the total mass of air enters the house, and the mass of air when the temperature decreases to -10 ⁰C are $101.32\text{ kg}$ .