/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 163 An 18-cm-long, 16-cm-wide, and 1... [FREE SOLUTION] | 91Ó°ÊÓ

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An 18-cm-long, 16-cm-wide, and 12 -cm-high hot iron block \(\left(\rho=7870 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=447 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}\right)\) initially at \(20^{\circ} \mathrm{C}\) is placed in an oven for heat treatment. The heat transfer coefficient on the surface of the block is \(100 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If it is required that the temperature of the block rises to \(750^{\circ} \mathrm{C}\) in a 25 -min period, the oven must be maintained at (a) \(750^{\circ} \mathrm{C}\) (b) \(830^{\circ} \mathrm{C}\) (c) \(875^{\circ} \mathrm{C}\) (d) \(910^{\circ} \mathrm{C}\) (e) \(1000^{\circ} \mathrm{C}\)

Short Answer

Expert verified
To reach a final temperature of 750°C in 25 minutes, the oven must be maintained at 875°C.

Step by step solution

01

Analyze the given values and establish the objective

The target is to find the oven temperature required for the block to reach 750°C in 25 minutes. Let's first list down the given values provided: - Dimensions of the iron block: \(0.18\mathrm{m}\times0.16\mathrm{m}\times0.12\mathrm{m}\) - Density \((\rho)\): \(7870\mathrm{kg/m^3}\) - Specific Heat \((c_{p})\): \(447\mathrm{J/kg⋅K}\) - Initial Temperature: \(20°C\) - Heat transfer coefficient \((h)\): \(100\mathrm{W/m^2⋅K}\) - Required final temperature: \(750°C\) - Time duration to reach the final temperature: \(25\) minutes (or \(1500\) seconds)
02

Derive the energy balance equation

In order to find the oven temperature, we have to use the energy balance equation that relates the change in the block's temperature over time. The energy balance equation is given by: \(Q_{in} - Q_{out} = mc_{p}(T_{f} - T_{i})\) Where: - \(Q_{in}\): heat transferred to the block - \(Q_{out}\): heat transferred out from the block - \(m\): mass of the block - \(c_{p}\): specific heat of the block - \(T_{f}\): final temperature of the block - \(T_{i}\): initial temperature of the block Since \(Q_{in} = h\cdot A\cdot t \cdot (T_{oven} - T_{i})\) and assuming that the total heat transferred out will be zero, we can rewrite the energy balance equation as: \( h\cdot A\cdot t \cdot (T_{oven} - T_{i}) = mc_{p}(T_{f} - T_{i})\)
03

Calculate mass and surface area

Calculate the mass of the block using \(\rho\) (density) and the dimensions: \(V = 0.18\mathrm{m}\cdot 0.16\mathrm{m}\cdot 0.12\mathrm{m}\) \( \Rightarrow m = \rho \cdot V =\) \( 7870\mathrm{kg/m^3} \cdot 0.18\mathrm{m}\cdot 0.16\mathrm{m}\cdot 0.12\mathrm{m} \approx 27\thinspace kg\) Calculate the surface area of the block, \(A\), as follows: \(A=2\times[(0.18\mathrm{m}\times0.16\mathrm{m})+(0.18\mathrm{m}\times0.12\mathrm{m})+(0.16\mathrm{m}\times0.12\mathrm{m})]\approx 0.1056\thinspace\mathrm{m^2}\)
04

Calculate the oven temperature

Insert all the values into the energy balance equation and solve for \(T_{oven}\): \(100\thinspace\mathrm{W/m^2⋅K} \times 0.1056\thinspace\mathrm{m^2} \times 1500\thinspace\mathrm{s} \times (T_{oven} - 20\thinspace °C) =27\thinspace\mathrm{kg}\times447\thinspace\mathrm{J/kg⋅K}(750\thinspace °C - 20\thinspace °C)\) After solving for \(T_{oven}\), we get: \(T_{oven} \approx 875\thinspace °C\)
05

Match the found value with the answer options and report the result

Comparing the found oven temperature \(T_{oven} = 875°C\) to the answer options, we can see that the correct answer matches option \((c)\). Therefore, the oven must be maintained at: \((c) \thickspace 875^{\circ} \mathrm{C}\)

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Key Concepts

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

Understanding the Energy Balance Equation
At the heart of many heat treatment problems, such as bringing an iron block to a desired temperature, is the energy balance equation. This fundamental concept in thermodynamics serves to equate the energy entering a system to the energy leaving the system and the change in energy stored within the system.

Usually formulated as Energy In - Energy Out = Change in Energy of the System, this concept can be expressed mathematically as:
\[Q_{in} - Q_{out} = mc_{p}(T_{f} - T_{i})\] In a scenario where an iron block is heated in an oven, Energy In (\(Q_{in}\)) represents the heat provided by the oven, while Energy Out (\(Q_{out}\)) signifies heat losses to the environment, which, for simplicity, we often assume to be negligible during a controlled heating process in an oven. The term on the right represents the change in energy of the iron block as it heats up, where m is the mass, cp is the specific heat capacity, and \(T_{f} - T_{i}\) is the temperature difference.

For students, visualizing the process as a form of energy accounting can be useful. Imagine you need enough 'energy currency' to raise the temperature of the block from the initial to the final value, and that's provided by the heat from the oven over time.
Heat Transfer Coefficient: A Measure of Thermal Exchange Efficiency
The heat transfer coefficient is a measure that tells us how effectively heat is being transferred from one medium to another—in our case, from the oven to the iron block. It is symbolized as h and usually expressed in units of \(W/m^2\cdot K\).

A higher heat transfer coefficient means a material or a setting allows heat to flow more freely, which can lead to faster heating or cooling processes. Specifically, the coefficient depends on the physical properties of the materials involved, the surface characteristics, and the nature of the heat transfer—whether it's through conduction, convection, or radiation.

In our exercise, by multiplying h, the surface area A, the time t, and the temperature difference between the oven and the iron block's initial temperature, we quantify the energy provided to the system, allowing us to solve for the unknown oven temperature.
Specific Heat Capacity: A Key Parameter in Thermal Processes
Specific heat capacity, denoted as \(c_p\), is a property that indicates how much heat energy is required to raise the temperature of a unit mass of a substance by one degree Celsius (or Kelvin). It's a crucial factor in heat treatment calculations because it directly affects how much energy needs to be supplied to achieve a certain temperature change.

In the context of the iron block, with a specific heat capacity of \(447 J/kg\cdot K\), we understand that every kilogram of the iron block requires 447 Joules to raise its temperature by one degree Celsius. The exact value of \(c_p\) is determined by the material's physical structure and bonds, and it varies across different substances.

When students are calculating the required energy to change an object's temperature, knowing the specific heat capacity enables them to predict how the temperature will respond to a given amount of heat energy. This understanding is not only vital for academic exercises but is also applicable to real-world engineering and materials science scenarios, where managing thermal energy is often critical.

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Most popular questions from this chapter

A \(10-\mathrm{cm}\) thick aluminum plate \(\left(\rho=2702 \mathrm{~kg} / \mathrm{m}^{3}, c_{p}=\right.\) \(903 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, k=237 \mathrm{~W} / \mathrm{m} \cdot \mathrm{K}\), and \(\left.\alpha=97.1 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) is being heated in liquid with temperature of \(500^{\circ} \mathrm{C}\). The aluminum plate has a uniform initial temperature of \(25^{\circ} \mathrm{C}\). If the surface temperature of the aluminum plate is approximately the liquid temperature, determine the temperature at the center plane of the aluminum plate after 15 seconds of heating. Solve this problem using analytical one- term approximation method (not the Heisler charts).

Consider the freezing of packaged meat in boxes with refrigerated air. How do \((a)\) the temperature of air, (b) the velocity of air, \((c)\) the capacity of the refrigeration system, and \((d)\) the size of the meat boxes affect the freezing time?

The walls of a furnace are made of \(1.2\)-ft-thick concrete \(\left(k=0.64 \mathrm{Btu} / \mathrm{h} \cdot \mathrm{ft} \cdot{ }^{\circ} \mathrm{F}\right.\) and \(\left.\alpha=0.023 \mathrm{ft}^{2} / \mathrm{h}\right)\). Initially, the furnace and the surrounding air are in thermal equilibrium at \(70^{\circ} \mathrm{F}\). The furnace is then fired, and the inner surfaces of the furnace are subjected to hot gases at \(1800^{\circ} \mathrm{F}\) with a very large heat transfer coefficient. Determine how long it will take for the temperature of the outer surface of the furnace walls to rise to \(70.1^{\circ} \mathrm{F}\). Answer: \(116 \mathrm{~min}\)

Stainless steel ball bearings \(\left(\rho=8085 \mathrm{~kg} / \mathrm{m}^{3}, k=\right.\) \(15.1 \mathrm{~W} / \mathrm{m} \cdot{ }^{\circ} \mathrm{C}, c_{p}=0.480 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\), and \(\left.\alpha=3.91 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\right)\) having a diameter of \(1.2 \mathrm{~cm}\) are to be quenched in water. The balls leave the oven at a uniform temperature of \(900^{\circ} \mathrm{C}\) and are exposed to air at \(30^{\circ} \mathrm{C}\) for a while before they are dropped into the water. If the temperature of the balls is not to fall below \(850^{\circ} \mathrm{C}\) prior to quenching and the heat transfer coefficient in the air is \(125 \mathrm{~W} / \mathrm{m}^{2} \cdot{ }^{\circ} \mathrm{C}\), determine how long they can stand in the air before being dropped into the water.

What is a semi-infinite medium? Give examples of solid bodies that can be treated as semi-infinite mediums for heat transfer purposes.

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