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

You are asked to design a cylindrical steel rod 50.0 cm long, with a circular cross section, that will conduct 190.0 J/s from a furnace at 400.0\(^\circ\)C to a container of boiling water under 1 atmosphere. What must the rod’s diameter be?

Short Answer

Expert verified
The rod's diameter must be approximately 2.13 cm.

Step by step solution

01

Understand the Problem

We need to find the diameter of a cylindrical steel rod that conducts heat at a rate of 190.0 J/s from a furnace at 400.0\(^\circ\)C to boiling water at 100.0\(^\circ\)C. The rod is 50.0 cm long, and we will use the formula for heat conduction.
02

Use the Heat Conduction Formula

The heat conduction formula is \( Q/t = \frac{kA(T_1 - T_2)}{L} \), where \( Q/t \) is the heat transfer rate (190.0 J/s), \( k \) is the thermal conductivity of steel, \( A \) is the cross-sectional area, \( T_1 \) and \( T_2 \) are the temperatures of the furnace and water, respectively, and \( L \) is the length of the rod.
03

Insert Known Values

Insert the known values into the equation: \( 190.0 = \frac{kA(400.0 - 100.0)}{0.5} \). Here, \( k \) for steel is approximately 50.2 W/m·K. Convert 50.0 cm to meters as 0.5 m.
04

Solve for Cross-Sectional Area

Rearrange the formula to solve for the cross-sectional area \( A \): \( A = \frac{(190.0 \times 0.5)}{(50.2 \times 300)} \). Calculate \( A \).
05

Calculate the Diameter

The cross-sectional area \( A \) of a circle is given by \( A = \pi (d/2)^2 \). Solve for the diameter \( d \) by rearranging the formula: \( d = 2 \times \sqrt{\frac{A}{\pi}} \). Substitute the value of \( A \) from the previous step to find \( d \).

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.

Thermal Conductivity
Thermal conductivity is a measure of a material's ability to conduct heat. It indicates how easily heat can pass through a material.
The higher the thermal conductivity, the better the material is at conducting heat.
In the context of a cylindrical steel rod, it is represented as the constant \( k \) in the heat conduction formula.Understanding this concept is crucial for applications where efficient heat transfer is required.
  • Steel has a moderate thermal conductivity, meaning it is reasonably good at transferring heat.
  • It is measured in watts per meter per kelvin (\( W/m \cdot K \)). In this exercise, steel's thermal conductivity is given as 50.2 \( W/m \cdot K \).
This value helps determine how much heat will be conducted over a given length of time.
Cylindrical Steel Rod
A cylindrical steel rod is a long, circular object often used for structural and mechanical purposes.
In heat conduction problems, its shape and material are significant factors.
The length of the rod influences how much heat can be transferred from one end to the other. In real-life applications, these rods are commonly used in heating systems and engineering projects.
  • The rod in this exercise is described as being 50.0 cm long, which should be converted to meters as 0.5 m for calculation purposes.
  • The rod's material, steel, affects both its mechanical properties and its thermal performance.
A solid understanding of these aspects is essential for effectively employing these rods in practical situations.
Cross-Sectional Area
The cross-sectional area of a rod is the area of its circular face that is perpendicular to its length.
It plays a crucial role in determining how much heat it can conduct.
The larger the cross-sectional area, the more heat it can carry from one end to the other.The formula for calculating the area is crucial:
  • For the circular cross section of a cylindrical rod, the area \( A \) is calculated as \( \pi \left( \frac{d}{2} \right)^2 \), where \( d \) is the diameter of the rod.
In heat conduction calculations, this area determines how much thermal energy is transferred over the rod's length and impacts the overall efficiency of the heat transfer process.
Temperature Difference
Temperature difference is defined as the difference in temperature between two points.
For heat conduction, it greatly influences the rate of thermal energy transfer.
The greater the difference in temperature, the higher the rate of heat transfer.This parameter is represented as \( T_1 - T_2 \) in the formula, where \( T_1 \) is the temperature of the hot side (furnace) and \( T_2 \) is the cooler side (boiling water):
  • In the given exercise, the temperature difference is 400.0°C - 100.0°C = 300.0°C.
  • This difference drives the heat transfer process down the steel rod.
Understanding the impact of temperature difference is essential in designing systems that efficiently manage heat flow.
Circular Cross Section
A circular cross section refers to the round shape of the rod's face when sliced perpendicular to its length.
This shape is common in cylindrical objects and is significant in calculations of area.This simple geometric property simplifies many engineering equations:
  • A circle's properties, such as its constant radius across all angles, facilitate straightforward calculations.
  • This shape helps in evenly distributing stress and avoiding sharp or weak points, a reason why it's favored in construction and engineering.
In the context of the given exercise, the circular cross section allows for the application of the formula \( A = \pi (d/2)^2 \). It is used to compute the area needed for heat conduction.

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

(a) Calculate the one temperature at which Fahrenheit and Celsius thermometers agree with each other. (b) Calculate the one temperature at which Fahrenheit and Kelvin thermometers agree with each other.

You have probably seen people jogging in extremely hot weather. There are good reasons not to do this! When jogging strenuously, an average runner of mass 68 kg and surface area 1.85 m\(^2\) produces energy at a rate of up to 1300 W, 80% of which is converted to heat. The jogger radiates heat but actually absorbs more from the hot air than he radiates away. At such high levels of activity, the skin's temperature can be elevated to around 33\(^\circ\)C instead of the usual 30\(^\circ\)C. (Ignore conduction, which would bring even more heat into his body.) The only way for the body to get rid of this extra heat is by evaporating water (sweating). (a) How much heat per second is produced just by the act of jogging? (b) How much net heat per second does the runner gain just from radiation if the air temperature is 40.0\(^\circ\)C (104\(^\circ\)F)? (Remember: He radiates out, but the environment radiates back in.) (c) What is the total amount of excess heat this runner's body must get rid of per second? (d) How much water must his body evaporate every minute due to his activity? The heat of vaporization of water at body temperature is \(2.42 \times 10{^6} J/kg\). (e) How many 750-mL bottles of water must he drink after (or preferably before!) jogging for a half hour? Recall that a liter of water has a mass of 1.0 kg.

CP While painting the top of an antenna 225 m in height, a worker accidentally lets a 1.00-L water bottle fall from his lunchbox. The bottle lands in some bushes at ground level and does not break. If a quantity of heat equal to the magnitude of the change in mechanical energy of the water goes into the water, what is its increase in temperature?

A laboratory technician drops a 0.0850-kg sample of unknown solid material, at 100.0\(^\circ\)C, into a calorimeter. The calorimeter can, initially at 19.0\(^\circ\)C, is made of 0.150 kg of copper and contains 0.200 kg of water. The final temperature of the calorimeter can and contents is 26.1\(^\circ\)C. Compute the specific heat of the sample.

BIO Temperatures in Biomedicine. (a) Normal body temperature. The average normal body temperature measured in the mouth is 310 K. What would Celsius and Fahrenheit thermometers read for this temperature? (b) Elevated body temperature. During very vigorous exercise, the body’s temperature can go as high as 40\(^\circ\)C. What would Kelvin and Fahrenheit thermometers read for this temperature? (c) Temperature difference in the body. The surface temperature of the body is normally about 7 C\(^\circ\) lower than the internal temperature. Express this temperature difference in kelvins and in Fahrenheit degrees. (d) Blood storage. Blood stored at 4.0\(^\circ\)C lasts safely for about 3 weeks, whereas blood stored at -160\(^\circ\)C lasts for 5 years. Express both temperatures on the Fahrenheit and Kelvin scales. (e) Heat stroke. If the body’s temperature is above 105\(^\circ\)F for a prolonged period, heat stroke can result. Express this temperature on the Celsius and Kelvin scales.

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