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**1.3 Temperature Increase due to Incandescent Lighting**

When energy is added to a fluid, the temperature of the fluid increases. An equation describing this phenomenon is:

\[ Q = M C_p \Delta T \]

Where:
- [tex]\( Q \)[/tex] is the amount of energy added (joules)
- [tex]\( M \)[/tex] is the mass of the fluid (kg)
- [tex]\( C_p \)[/tex] is the heat capacity of the fluid (joules/kg K)
- [tex]\(\Delta T\)[/tex] is the change in temperature (K, or °C)

A garage (24 ft × 24 ft × 10 ft) is illuminated by six 60 W incandescent bulbs. It is estimated that 90% of the energy from an incandescent bulb is dissipated as heat. If the bulbs are on for 3 hours, how much would the temperature in the garage increase because of the light bulbs (assuming no energy losses)?

Complete an Excel sheet like the one illustrated in Figure 1.71 to answer this question.

Potentially useful information:
- Air Density (approximate): [tex]1.2 \, \text{kg/m}^3[/tex]
- Air Heat Capacity (approximate): 1000 joules/kg K
- [tex]3.28 \, \text{ft} = 1 \, \text{m}[/tex]

Answer :

To solve this problem, we need to calculate how much the temperature in the garage increases when it's illuminated by six 60-watt incandescent bulbs for three hours. Here's a step-by-step guide:

1. Calculate the Volume of the Garage:
- The dimensions of the garage are given in feet: 24 ft by 24 ft by 10 ft.
- Convert these dimensions from feet to meters. We use the conversion factor [tex]\(1 \text{ ft} = 0.3048 \text{ m}\)[/tex].
- Length = [tex]\(24 \times 0.3048 = 7.3152 \text{ m}\)[/tex]
- Width = [tex]\(24 \times 0.3048 = 7.3152 \text{ m}\)[/tex]
- Height = [tex]\(10 \times 0.3048 = 3.048 \text{ m}\)[/tex]
- The volume of the garage (in cubic meters) is then:
[tex]\[
\text{Volume} = 7.3152 \times 7.3152 \times 3.048 = 163.105 \text{ m}^3
\][/tex]

2. Calculate the Total Energy Dissipated as Heat by the Bulbs:
- Each bulb has a power of 60 watts, and it's given that 90% of this energy is converted to heat.
- There are six bulbs, and they are on for 3 hours. Convert hours to seconds: [tex]\(3 \times 3600 = 10800 \text{ seconds}\)[/tex].
- Total energy (in joules) is calculated by:
[tex]\[
\text{Total Energy} = 6 \times 60 \times 0.90 \times 10800 = 3499200 \text{ joules}
\][/tex]

3. Calculate the Mass of the Air in the Garage:
- Use the approximate air density of [tex]\(1.2 \text{ kg/m}^3\)[/tex].
- The mass of the air is:
[tex]\[
\text{Mass} = 1.2 \times 163.105 = 195.726 \text{ kg}
\][/tex]

4. Calculate the Temperature Increase (ΔT):
- The specific heat capacity of air is given as [tex]\(1000 \text{ J/kg} \cdot \text{K}\)[/tex].
- We use the formula [tex]\(Q = M \cdot C_p \cdot \Delta T\)[/tex] to find the temperature increase.
- Solving for [tex]\(\Delta T\)[/tex], we get:
[tex]\[
\Delta T = \frac{3499200}{195.726 \times 1000} = 17.878 \text{ °C}
\][/tex]

Thus, the temperature in the garage increases by approximately [tex]\(17.88 \text{ °C}\)[/tex] after the bulbs are on for 3 hours.

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