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Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a 60 -W lightbulb. You will need a thermometer, which can be purchased in a hardware store, and metal glue. You will also need a piece of string and a ruler to calculate the surface area of the lightbulb. First, measure the air temperature in the room, and then glue the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. Turn the light on and wait until the temperature reading stabilizes. The temperature reading will give the surface temperature of the lightbulb. Assuming 10 percent of the rated power of the bulb is converted to light and is transmitted by the glass, calculate the heat transfer coefficient from Newton's law of cooling.

Short Answer

Expert verified
Short Answer: To find the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces, you need to follow these steps: 1. Measure the air temperature (T_air) and the lightbulb's surface temperature (T_surface) with a thermometer. 2. Determine the power output in the form of heat (P_heat) by calculating 90 percent of the bulb's rated power (e.g., for a 60W bulb, P_heat = 54 watts). 3. Calculate the surface area (A) of the lightbulb, approximating it as a sphere. 4. Use Newton's law of cooling formula, Q = h * A * (T_surface - T_air), and the power output in the form of heat (P_heat) to calculate the heat transfer coefficient (h). Once you've calculated the value of h, you've found the combined heat transfer coefficient between the incandescent lightbulb and the surrounding air and surfaces.

Step by step solution

01

Measure the air temperature

Using a thermometer, measure the temperature of the surrounding air in the room. Record this value as T_air.
02

Attach the thermocouple to the lightbulb

Apply metal glue to the tip of the thermocouple wire of the thermometer and attach it to the glass surface of the lightbulb. Wait for the glue to cure according to the manufacturer's instructions.
03

Turn on the light and measure the surface temperature

After the glue is cured, turn on the light and wait until the temperature reading on the thermometer stabilizes. This value represents the surface temperature of the lightbulb. Record this value as T_surface.
04

Determine the power output in the form of heat

Since only 10 percent of the bulb's rated power (60 watts) is converted to light and is transmitted by the glass, we can determine the power output in the form of heat by calculating 90 percent of 60 watts. P_heat = (90/100) * 60 = 54 watts
05

Calculate the surface area of the lightbulb

Using the piece of string and ruler, measure the circumference and diameter of the lightbulb. The lightbulb can be approximated as a sphere. Calculate the surface area (A) of the lightbulb using the formula for the surface area of a sphere: A = 4 * pi * (radius^2)
06

Calculate the heat transfer coefficient using Newton's law of cooling

Newton's law of cooling states that the heat transfer rate (Q) is proportional to the temperature difference between the surface (T_surface) and the surrounding air (T_air), with the proportionality constant being the heat transfer coefficient (h) and the surface area (A): Q = h * A * (T_surface - T_air) We have calculated the power output in the form of heat (P_heat) and the surface area (A), so we can solve for the heat transfer coefficient (h): h = Q / (A * (T_surface - T_air)) Since Q = P_heat = 54 watts, and we already determined A and measured T_air and T_surface: h = 54 / (A * (T_surface - T_air)) Once you've calculated h, you've found the combined heat transfer coefficient between the incandescent lightbulb and the surrounding air and surfaces.

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Most popular questions from this chapter

A person's head can be approximated as a \(25-\mathrm{cm}\) diameter sphere at \(35^{\circ} \mathrm{C}\) with an emissivity of \(0.95\). Heat is lost from the head to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of $11 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\( and by radiation to the surrounding surfaces at \)10^{\circ} \mathrm{C}$. Disregarding the neck, determine the total rate of heat loss from the head. (a) \(22 \mathrm{~W}\) (b) \(27 \mathrm{~W}\) (c) \(49 \mathrm{~W}\) (d) \(172 \mathrm{~W}\) (e) \(249 \mathrm{~W}\)

Consider heat loss through two walls of a house on a winter night. The walls are identical except that one of them has a tightly fit glass window. Through which wall will the house lose more heat? Explain.

The boiling temperature of nitrogen at atmospheric pressure at sea level $(1 \mathrm{~atm})\( is \)-196^{\circ} \mathrm{C}$. Therefore, nitrogen is commonly used in low-temperature scientific studies since the temperature of liquid nitrogen in a tank open to the atmosphere remains constant at $-196^{\circ} \mathrm{C}$ until the liquid nitrogen in the tank is depleted. Any heat transfer to the tank results in the evaporation of some liquid nitrogen, which has a heat of vaporization of \(198 \mathrm{~kJ} / \mathrm{kg}\) and a density of \(810 \mathrm{~kg} / \mathrm{m}^{3}\) at \(1 \mathrm{~atm}\). Consider a 4-m-diameter spherical tank initially filled with liquid nitrogen at \(1 \mathrm{~atm}\) and \(-196^{\circ} \mathrm{C}\). The tank is exposed to \(20^{\circ} \mathrm{C}\) ambient air with a heat transfer coefficient of $25 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$. The temperature of the thin- shelled spherical tank is observed to be almost the same as the temperature of the nitrogen inside. Disregarding any radiation heat exchange, determine the rate of evaporation of the liquid nitrogen in the tank as a result of the heat transfer from the ambient air.

An ice skating rink is located in a building where the air is at $T_{\text {air }}=20^{\circ} \mathrm{C}\( and the walls are at \)T_{w}=25^{\circ} \mathrm{C}$. The convection heat transfer coefficient between the ice and the surrounding air is \(h=10 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). The emissivity of ice is \(\varepsilon=0.95\). The latent heat of fusion of ice is \(h_{i f}=333.7 \mathrm{~kJ} / \mathrm{kg}\), and its density is $920 \mathrm{~kg} / \mathrm{m}^{3}$. (a) Calculate the refrigeration load of the system necessary to maintain the ice at \(T_{s}=0^{\circ} \mathrm{C}\) for an ice rink of \(12 \mathrm{~m}\) by \(40 \mathrm{~m}\). (b) How long would it take to melt \(\delta=3 \mathrm{~mm}\) of ice from the surface of the rink if no cooling is supplied and the surface is considered insulated on the back side?

A 30 -cm-diameter black ball at \(120^{\circ} \mathrm{C}\) is suspended in air. It is losing heat to the surrounding air at \(25^{\circ} \mathrm{C}\) by convection with a heat transfer coefficient of $12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}$ and by radiation to the surrounding surfaces at \(15^{\circ} \mathrm{C}\). The total rate of heat transfer from the black ball is (a) \(322 \mathrm{~W}\) (b) \(595 \mathrm{~W}\) (c) \(234 \mathrm{~W}\) (d) \(472 \mathrm{~W}\) (e) \(2100 \mathrm{~W}\)

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