Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

A 70-cm-diameter flat black disk is placed at the center of the ceiling of a 1m×1m×1m black box. If the temperature of the box is 620C and the temperature of the disk is 27C, the rate of heat transfer by radiation between the interior of the box and the disk is (a) 2 kW (b) 5 kW (c) 8 kW (d) 11 kW (e) 14 kW

Short Answer

Expert verified
Answer: (b) 5 kW

Step by step solution

01

Convert temperatures to Kelvin

Before we start our calculations, we need to convert the temperatures from Celsius to Kelvin for both the box and the disk. To convert temperatures from Celsius to Kelvin, add 273.15 to the Celsius temperature: Box temperature: 620°C + 273.15 = 893.15 K Disk temperature: 27°C + 273.15 = 300.15 K
02

Calculate the area of the disk

We need to find the area of the flat black disk. Since it's a disk, its area is given by the formula A = πr² where r is the radius and A is the area. First, find the radius of the disk: The diameter is 70cm, so the radius is half of that which is 35 cm. To convert it to meters, divide by 100: 35cm / 100 = 0.35 m. Now calculate the area of the disk using the formula A = πr²: A = π(0.35)^2 = 0.3849 m²
03

Calculate the rate of heat transfer by radiation

To find the rate of heat transfer between the box and the disk, use the Stefan-Boltzmann Law formula: Q = Aσ(Tb^4 - Td^4) where Q is the rate of heat transfer, A is the area of the disk, σ is the Stefan-Boltzmann constant (5.67 x 10^-8 W/(m²K^4)), Tb is the temperature of the box, and Td is the temperature of the disk. Now plug in the values that we have: Q = (0.3849 m²)(5.67 x 10^-8 W/(m²K^4))((893.15 K)^4 - (300.15 K)^4) ≈ 5070 W Now convert the rate of heat transfer to kW by dividing by 1000: Q = 5070 / 1000 ≈ 5.07 kW
04

Choose the correct answer

Now look at the given options and choose the closest value to our calculated heat transfer rate (5.07 kW): (a) 2 kW (b) 5 kW (c) 8 kW (d) 11 kW (e) 14 kW Thus, the correct answer is (b) 5 kW.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Stefan-Boltzmann Law
The Stefan-Boltzmann Law is a fundamental principle in thermodynamics relating to heat transfer by radiation. It specifies how the thermal radiation emitted by a black body is proportional to the fourth power of its absolute temperature. The law is mathematically expressed by the equation:
Q=Aσ(Tb4Td4)
where:Unknown environment 'itemize'
The equation tells us that the rate at which a body radiates energy increases rapidly as the temperature rises. For instance, doubling the temperature of an object increases its thermal radiation by a factor of 16, as this relationship is to the fourth power (24=16). Due to this, temperature differences play a significant role in the transfer of heat by radiation, which is crucial when calculating the energy exchange between objects at different temperatures, such as in our exercise with the disk and the box.
Temperature Conversion
In physics problems, it's essential to use the correct units. For temperature-related calculations, we often need to convert Celsius to Kelvin - the standard unit in the International System (SI) for thermodynamic temperature. The conversion is straightforward but crucially important for ensuring accurate calculations in thermodynamics. To convert from Celsius to Kelvin, we use the following relationship:
K=\degreeC+273.15
where:Unknown environment 'itemize'
Adding 273.15 to the Celsius temperature gives us the temperature in Kelvin necessary for computations involving the Stefan-Boltzmann Law, as seen in our example exercise. Precision is key, and overlooking this step can lead to incorrect answers. This simple yet essential conversion allows us to accurately assess how heat is transferred by radiation between objects at different temperatures, like our flat black disk and the interior of the box in the given problem.
Area Calculation of a Disk
The determination of an object's surface area is often required when dealing with heat transfer calculations. For objects like our flat black disk, we use area calculation formulas unique to their geometrical shape.
The area (A) of a disk is calculated by the formula: A=πr2
where:Unknown environment 'itemize'
To calculate the area in our example, we first need to determine the radius, which is half of the diameter. After finding the radius, we apply the formula. Although the calculation itself is straightforward, precision is necessary—it's about using the correct units and calculating to adequate significance. Here, ensuring the radius is in meters is key, as the Stefan-Boltzmann Law uses SI units for area. Therefore, if the radius is initially given in centimeters, it must be converted to meters before applying the area formula. Understanding how to calculate the area correctly enables us to quantify the heat transfer across the disk in our sample problem.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Consider the two parallel coaxial disks of diameters a and b, shown in Fig. P13-131. For this geometry, the view factor from the smaller disk to the larger disk can be calculated from Missing \left or extra \right where, A=a/2L,B=b/2L, and C=1+[(1+A2)/B2]. The diameter, emissivity and temperature are 20 cm,0.60, and 600C, respectively, for disk a, and 40 cm,0.80 and 200C for disk b. The distance between the two disks is L=10 cm. (a) Calculate Fab and Fba. (b) Calculate the net rate of radiation heat exchange between disks a and b in steady operation. (c) Suppose another (infinitely) large disk c, of negligible thickness and ε=0.7, is inserted between disks a and b such that it is parallel and equidistant to both disks. Calculate the net rate of radiation heat exchange between disks a and c and disks c and b in steady operation.

A 2-m-internal-diameter double-walled spherical tank is used to store iced water at 0C. Each wall is 0.5 cm thick, and the 1.5-cm-thick air space between the two walls of the tank is evacuated in order to minimize heat transfer. The surfaces surrounding the evacuated space are polished so that each surface has an emissivity of 0.15. The temperature of the outer wall of the tank is measured to be 20C. Assuming the inner wall of the steel tank to be at 0C, determine (a) the rate of heat transfer to the iced water in the tank and (b) the amount of ice at 0C that melts during a 24-h period.

Two parallel concentric disks, 20 cm and 40 cm in diameter, are separated by a distance of 10 cm. The smaller disk (ε=0.80) is at a temperature of 300C. The larger disk (ε=0.60) is at a temperature of 800C. (a) Calculate the radiation view factors. (b) Determine the rate of radiation heat exchange between the two disks. (c) Suppose that the space between the two disks is completely surrounded by a reflective surface. Estimate the rate of radiation heat exchange between the two disks.

What is the crossed-strings method? For what kind of geometries is the crossed-strings method applicable?

What is operative temperature? How is it related to the mean ambient and radiant temperatures? How does it differ from effective temperature?

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free