Chapter 14, Problem 56GP

(a)

To determine

To Calculate: The rate at which heat must be supplied to the house.

Answer to Problem 56GP

The rate at which heat must be supplied to the house to maintain the its inside temperature is $\text{1}{\text{.6×10}}^{\text{5}}\text{W}$ .

Explanation of Solution

Given:

Walls’s thickness is 17.5 cm.

Area of the wall, $\text{A}\text{=}\text{410}{\text{m}}^{\text{2}}$

Thickness of roof, $\text{t}\text{=}\text{6}\text{.5cm}$

Area of the roof, ${\text{A}}_{\text{roof}}\text{=}\text{280}{\text{cm}}^{\text{2}}$

Uncovered window thickness, $t_{window}=0.65\text{cm}$

Total area, $A_{total}=33{\text{m}}^{\text{2}}$

Temperatures are $\left(\text{23}°\text{C}\right)\text{and}\left(\text{-10}°\text{C}\right)$ .

Formula used:

The heat transfer rate by the conduction is calculated as,

  $\frac{Q_{conducntion}}{t}=\left[{\left(\frac{kA}{l}\right)}_{walls}+{\left(\frac{kA}{l}\right)}_{roof}+{\left(\frac{kA}{l}\right)}_{windows}\right]\left(T_1-T_2\right)$

Where, k is the constant value, A is the area and $l$ is the respective thickness.

Calculation:

The heat transfer for all three areas such as walls, roof and windows and each area have the same temperature difference. Therefore, the rate is calculated by using the given formula such that,

  $\begin{array}{l}\frac{Q_{conducntion}}{t}=\left[{\left(\frac{kA}{l}\right)}_{walls}+{\left(\frac{kA}{l}\right)}_{roof}+{\left(\frac{kA}{l}\right)}_{windows}\right]\left(T_1-T_2\right)\\ =\left[\left(\frac{\left(\text{0}\text{.023}\text{J/s}\cdot \text{m}\cdot °\text{C}\right)\text{×}\left(\text{410}{\text{m}}^{\text{2}}\right)}{\text{1}{\text{.75×10}}^{\text{-1}}\text{m}}\right)\text{+}\left(\frac{\left(\text{0}\text{.12}\text{J/s}\cdot \text{m}\cdot °\text{C}\right)\text{×}\left(\text{280}{\text{m}}^{\text{2}}\right)}{\text{6}{\text{.5×10}}^{\text{-2}}\text{m}}\right)\text{+}\left(\frac{\left(\text{0}\text{.84}\text{J/s}\cdot \text{m}\cdot °\text{C}\right)\text{×}\left(\text{33}{\text{m}}^{\text{2}}\right)}{\text{6}{\text{.5×10}}^{\text{-3}}\text{m}}\right)\right]\left(\text{33}°\text{C}\right)\\ \text{=}\left[\text{1}{\text{.596×10}}^{\text{5}}\text{W}\right]\approx \text{1}{\text{.6×10}}^{\text{5}}\text{W}\end{array}$

Conclusion:

The rate at which heat must be supplied to the house is $\text{1}{\text{.6×10}}^{\text{5}}\text{W}$ .

(b)

To determine

To Calculate: The heat the amount of heat that must be supplied to raise the temperature.

Answer to Problem 56GP

  $\text{2}{\text{.4×10}}^{\text{8}}\text{J}$

Explanation of Solution

Given:

Walls’s thickness is 17.5 cm.

Area of the wall, $\text{A}\text{=}\text{410}{\text{m}}^{\text{2}}$

Thickness of roof, $\text{t}\text{=}\text{6}\text{.5cm}$

Area of the roof, ${\text{A}}_{\text{roof}}\text{=}\text{280}{\text{cm}}^{\text{2}}$

Uncovered window thickness, $t_{window}=0.65\text{cm}$

Total area, $A_{total}=33{\text{m}}^{\text{2}}$

Temperatures are $\left(\text{23}°\text{C}\right)\text{and}\left(\text{-10}°\text{C}\right)$ .

Time period is 30 minutes.

Volume, ${\text{V=750m}}^{\text{3}}$

Formula used:

The amount of heat is calculated as,

  ${\text{Q}}_{Added}={\text{Q}}_{\begin{array}{l}raise\\ temperature\end{array}}+{\text{Q}}_{conduction}$

Where,

  ${\text{Q}}_{\begin{array}{l}raise\\ temperature\end{array}}$ is the amount of heat in raised temperature case, ${\text{Q}}_{conduction}$ is the heat amount in the conduction case.

Calculation:

The air is heated, and the energy is lost by conduction by adding the energy. Therefore, the required heat for raise the temperature is given as

  ${\text{Q}}_{\begin{array}{l}raise\\ temperature\end{array}}=m_{air}{c}_{air}{\left(\Delta T\right)}_{\text{warming}}$

The heat loss in conduction is directly proportional to the temperature difference between inside and outside which is given as, $\left(\text{23}°\text{C}\right)\text{and}\left(\text{-10}°\text{C}\right)$

So, the average temperature is $\left(\text{26}{\text{.5}}^{\text{0}}\text{C}\right)$ and the amount of heart which is required to raise the temperature is,

  $\begin{array}{l}{\text{Q}}_{\text{Added}}{\text{=Q}}_{\begin{array}{l}\text{raise}\\ \text{temperature}\end{array}}{\text{+Q}}_{\text{conduction}}\\ {\text{=m}}_{\text{air}}{\text{c}}_{\text{air}}{\left(\text{ΔT}\right)}_{\text{warming}}\text{+}\left(\frac{{\text{Q}}_{\text{conducntion}}}{\text{t}}\right)\text{×}\left(\text{1800}\text{s}\right)\\ \text{=}\left(\text{1}\text{.29}{\text{kg/m}}^{\text{3}}\right)\text{×}\left(\text{750}{\text{m}}^{\text{3}}\right)\text{×}\left(\text{0}\text{.24}\text{kcal/kg}\cdot °\text{C}\right)\text{×}\left(\frac{\text{4186J}}{\text{kcal}}\right)\text{×}\left(\text{13}°\text{C}\right)\text{+}\left(\text{1}{\text{.596×10}}^{\text{5}}\text{J/s}\right)\text{×}\left(\frac{\text{26}\text{.5}°\text{C}}{\text{33}°\text{C}}\right)\left(\text{1800}\text{s}\right)\\ \text{=}\text{2}{\text{.4×10}}^{\text{8}}\text{J}\end{array}$

Conclusion:

The required heat is $\text{2}{\text{.4×10}}^{\text{8}}\text{J}$ .

(c)

To determine

To Calculate: The monthly cost for the maintenance.

Answer to Problem 56GP

  $\$681$

Explanation of Solution

Given:

Time period is 30 minutes.

Volume, ${\text{V=750m}}^{\text{3}}$

Heat combustion is $5.4\times 10^7\text{J/kg}$ .

The cost is $\$0.080\text{ per kg}$ .

Formula used:

The amount of heat is calculated as,

  ${\text{Q}}_{Added}={\text{Q}}_{\begin{array}{l}raise\\ temperature\end{array}}+{\text{Q}}_{conduction}$

Where, ${\text{Q}}_{\begin{array}{l}raise\\ temperature\end{array}}$ is the amount of heat to raise temperature and ${\text{Q}}_{conduction}$ is the heat amount in the conduction case.

Calculation:

The amount of heat for the given gas is

  $\begin{array}{l}\text{0}{\text{.9Q}}_{\text{gas}}\text{=}{\left(\frac{\text{Q}}{\text{t}}\right)}_{\text{conduction}}{\text{×t}}_{\text{month}}\\ {\text{Q}}_{\text{gas}}\text{=}\frac{\text{1}}{\text{0}\text{.9}}{\left(\frac{\text{Q}}{\text{t}}\right)}_{\text{conduction}}{\text{×t}}_{\text{month}}\\ \text{=}\frac{\text{1}}{\text{0}\text{.9}}\text{×}\left(\text{1}{\text{.596×10}}^{\text{5}}\text{J/s}\right)\text{×}\left(\text{30}\text{days}\right)\text{×}\left(\frac{\text{24}\text{hr}}{\text{1day}}\right)\text{×}\left(\frac{\text{3600}\text{s}}{\text{1hr}}\right)\\ \text{=}\text{4}{\text{.596×10}}^{\text{11}}\text{J}\end{array}$

The monthly cost is for a month in which there are 30 days are present, is

  $\begin{array}{l}\left(\text{4}{\text{.596×10}}^{\text{11}}\text{J}\right)\text{×}\left(\frac{\text{1kg}}{\text{5}{\text{.4×10}}^{\text{7}}\text{J}}\right)\left(\frac{\$0.08}{\text{kg}}\right)=\$681\end{array}$

Conclusion:

The monthly cost is $\$681$ .