Conduct this experiment to determine the combined heat transfer coefficient between an incandescent lightbulb and the surrounding air and surfaces using a \(60-\mathrm{W}\) lightbulb. You will need a thermometer, which can be purchased in a hardware store, and a 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.

Step by step solution

01

Measure the Air Temperature

Prepare a thermometer and read the surrounding air temperature. This value will be used to determine the temperature difference between the bulb and the surrounding air.

02

Attach the Thermometer to the Lightbulb

Using the metal glue, attach the tip of the thermocouple wire of the thermometer to the glass of the lightbulb. The goal is to monitor the surface temperature of the bulb once it is turned on.

03

Determine the Surface Area of the Lightbulb

Measure the circumference of the lightbulb using a piece of string and a ruler. An approximate surface area can be calculated by considering the lightbulb as a sphere with the measured circumference, \(C\). Compute the surface area \(A\) using the formula: \(A = 4 \pi r^2\), with \(r = \frac{C}{2\pi}\).

04

Turn on the Lightbulb and Take Temperature Measurements

Turn on the lightbulb and wait for the temperature reading on the thermometer to stabilize. This will give the surface temperature of the lightbulb. Note down the stabilized temperature reading.

05

Calculate the Heat Transfer Coefficient using Newton's Law of Cooling

According to the problem description, 10% of the rated power of the bulb is converted to light, so 90% of the power will be converted to heat. The power of the bulb is \(60 \mathrm{W}\), so the power converted to heat is \(0.9 \times 60 = 54 \mathrm{W}\). Newton's law of cooling states that the heat transfer rate \(Q\) is proportional to the temperature difference between the bulb and the surrounding air, i.e., \(Q = hA(T_b - T_a)\), where \(h\) is the heat transfer coefficient, \(A\) is the surface area of the bulb, \(T_b\) is the temperature of the bulb, and \(T_a\) is the air temperature. We already have the values for \(Q\), \(A\), \(T_b\), and \(T_a\). Thus, we can solve for \(h\) by re-arranging the formula: $$h = \frac{Q}{A(T_b - T_a)}$$ Plug in the values and calculate the heat transfer coefficient \(h\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Newton's Law of Cooling

Newton's Law of Cooling describes the rate at which an object cools down or warms up due to heat transfer. It's based on the simple idea that the rate of heat transfer is proportional to the difference in temperatures between the object and its surroundings. When a lightbulb heats up, it loses heat to the surrounding air at a rate that can be modeled using this principle.

The law can be represented mathematically by the equation: \[Q = hA(T_b - T_a)\], where:\[Q\] is the heat transfer rate, \[h\] is the heat transfer coefficient, \[A\] is the surface area of the object, \[T_b\] is the temperature of the object, and \[T_a\] is the temperature of the surrounding air. By rearranging this equation, we can solve for the heat transfer coefficient \(h\) when we know the other variables.

Thermocouple

A thermocouple is a sensor used for measuring temperature. It consists of two different types of metals, joined together at one end. When the junction of the two metals is heated or cooled, a voltage is created that can be correlated directly to the temperature. Thermocouples are widely used because of their wide range of temperatures, durability, and fast response to temperature changes.

For the experiment involving a lightbulb, the thermocouple is crucial for obtaining an accurate measurement of the bulb's surface temperature. The metal glue used ensures good thermal contact between the thermocouple and the lightbulb, which in turn helps us compute the heat transfer coefficient more precisely.

Conductive Heat Transfer

Conductive heat transfer is one of the three modes of heat transfer, which also include convection and radiation. It is the process of heat energy being transmitted through collisions between neighboring molecules in a substance. Conductivity depends on the material and affects how quickly heat can be transferred through it.

In the case of the lightbulb in our exercise, conductive heat transfer occurs between the thermocouple and the lightbulb. The metal glue plays a role in enhancing this conductive path by ensuring a good thermal connection between the two, allowing for an accurate measurement of temperature.

Surface Temperature Measurement

Surface temperature measurement is essential in various fields such as material science, engineering, and meteorology. In our scenario, measuring the temperature of a lightbulb's surface is critical to determining the heat transfer coefficient. Thermocouples are among the most common tools for this purpose due to their precise readings and fast response. The correct placement of the thermocouple on the bulb's surface ensures that the temperature measurement reflects the surface temperature without being influenced by the ambient air temperature or other factors. The ability to measure surface temperature accurately is crucial to applying Newton's Law of Cooling and calculating the heat transfer coefficient.