/*! 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 63 A \(100-\mathrm{kg}\) steel supp... [FREE SOLUTION] | 91Ó°ÊÓ

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A \(100-\mathrm{kg}\) steel support rod in a building has a length of \(2.0 \mathrm{~m}\) at a temperature of \(20^{\circ} \mathrm{C}\). The rod supports a hanging load of \(6000 \mathrm{~kg}\). Find (a) the work done on the rod as the temperature increases to \(40^{\circ} \mathrm{C}\), (b) the energy \(Q\) added to the rod (assume the specific heat of steel is the same as that for iron), and (c) the change in internal energy of the rod.

Short Answer

Expert verified
The work done on the rod as the temperature increases to 40^{\circ}C is 28.22 J, the energy added to the rod is 900000 J and the change in internal energy of the rod is 899971.78 J.

Step by step solution

01

Solution for Part (a): Work Done on the Rod

The formula for the work done is given by: \(W = F \cdot d\). The force \(F\) is equal to the weight of the load attached \(6000 \mathrm{~kg} \cdot 9.8 \mathrm{~m/s^2} = 58800 \mathrm{~N}\). To find \(d\), we need to calculate the change in length of the rod due to thermal expansion. This is given by: \(d = L \cdot \alpha \cdot \Delta T\). Given \(L = 2.0 \mathrm{~m}\), \(\Delta T = 40^{\circ}C - 20^{\circ}C = 20^{\circ}C \), and \(\alpha = 12 \times 10^{-6} K^{-1}\) (coefficient of linear expansion for steel), substituting these gives \(d = 0.00048 \mathrm{~m}\). Substituting \(F\) and \(d\) in \(W = F \cdot d\) gives \(W = 58800 \mathrm{~N} \cdot 0.00048 \mathrm{~m} = 28.22 \mathrm{~J}\). Hence, the work done on the rod as the temperature increases to 40^{\circ}C is 28.22 J.
02

Solution for Part (b): Energy Added to the Rod

The energy \(Q\) added is computed using the specific heat formula, \(Q = m \cdot c \cdot \Delta T\). Given that the specific heat \(c\) for iron (and in this case, for steel) is 450 J/kg.K, and the mass \(m\) of the rod is 100 kg, as well as the change in temperature \(\Delta T = 20^{\circ}C\), substituting these values into the formula gives \(Q = 100 \mathrm{~kg} \cdot 450 \mathrm{~J/kg.K} \cdot 20 K = 900000 J\). Therefore, the energy added to the rod is 900000 J.
03

Solution for Part (c): Change in Internal Energy

The first law of thermodynamics states that the change in internal energy \(\Delta U\) of a system is the sum of the heat added to the system \(Q\) and the work \(W\) done on the system. Mathematically, this is formulated as \(\Delta U = Q - W\). Using our values for \(Q = 900000 J\) and \(W = 28.22 J\), we find that \(\Delta U = 900000 J - 28.22 J = 899971.78 J\). Therefore, the change in internal energy of the rod is 899971.78 J.

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Key Concepts

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

Coefficient of Linear Expansion
Imagine a metal rod, seemingly rigid and unchanging. However, when temperatures change, so does the rod. This happens due to the coefficient of linear expansion, a value that quantifies the change in length of a material when its temperature is altered.

For instance, a steel rod at a comfy room temperature of 20°C might stretch just a tiny bit if the room gets warmer. This change can be predicted with remarkable precision using the formula, \(d = L \cdot \alpha \cdot \Delta T\), where \(L\) is the original length, \(\alpha\) is the coefficient of linear expansion for the material (for steel it's typically 12 x 10^-6 per Kelvin), and \(\Delta T\) is the change in temperature (in Kelvins or degrees Celsius, since they have the same increment size).

In the case of our textbook problem, the steel rod expands by \(0.00048 m\) when heated from 20°C to 40°C, precisely because of this phenomenon. This seemingly minor detail plays a pivotal role in engineering applications, such as the stability of structures and the design of components subjected to varying temperatures.
Specific Heat
Now let's turn up the heat, quite literally. Specific heat is like a thermal fingerprint of a material. It tells us how much energy you need to pump into a certain amount of mass to raise its temperature by one degree.

We use the equation \(Q = m \cdot c \cdot \Delta T\) to calculate the energy required, where \(Q\) is the energy added to the material, \(m\) is its mass, \(c\) is the specific heat, and \(\Delta T\) is the change in temperature. For a 100-kg steel rod, with the same specific heat as iron (450 J/kg.K), heating it up by 20°C necessitates a whopping 900,000 Joules of energy.

It's akin to running a microwave for quite a long stretch to warm your dinner, but in this case, it's heating the metal. This concept is crucial when we discuss energy transfer, efficiency in heating processes, and even in our daily lives, from cooking to climate control systems.
First Law of Thermodynamics
The First Law of Thermodynamics is the universe's way of telling us that there's no such thing as a free lunch. In scientific terms, it's all about the conservation of energy within a system. This law states that the change in internal energy of a system is equal to the heat added to it minus the work done by it.

Mathematically, it's expressed as \(\Delta U = Q - W\), where \(\Delta U\) is the change in internal energy, \(Q\) is the heat added to the system, and \(W\) is the work done by the system. Back to our steel rod in the building example: when it absorbs 900,000 J of heat and the work done on the rod is a lesser 28.22 J, the internal energy increases by nearly the total heat added, since the work done is a negligible subtraction.

This principle also underscores the nature of energy conservation. Whether it's a simple rod expanding with heat, a steam engine chugging along, or the cells in your body metabolizing nutrients, the First Law of Thermodynamics rules them all, ensuring energy within a closed system is never created or destroyed – only transformed.

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