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In a certain process, \(8.00\) kcal of heat is furnished to the system while the system does \(6.00 \mathrm{~kJ}\) of work. By how much does the internal energy of the system change during the process? Here \(8.00\) kcal is heat-in and \(6.00 \mathrm{~kJ}\) is work-out, both of which are positive. Consequently, \(\Delta Q=(8000\) cal \()(4.184 \mathrm{~J} / \mathrm{cal})=33.5 \mathrm{~kJ}\) and \(\Delta W=6.00 \mathrm{~kJ}\) Therefore, from the First Law \(\Delta Q=\Delta U+\Delta W\), $$\Delta U=\Delta Q-\Delta W=33.5 \mathrm{~kJ}-6.00 \mathrm{~kJ}=27.5 \mathrm{~kJ}$$

Short Answer

Expert verified
The internal energy change is 27.472 kJ.

Step by step solution

01

Understanding Given Values and Units

We have been given that heat furnished to the system is 8.00 kcal and work done by the system is 6.00 kJ. We need to convert these into consistent units for calculations, usually in joules or kilojoules.
02

Converting Heat to Joules

Convert the heat from kilocalories to kilojoules, knowing that 1 kcal equals 4184 J. Thus, 8.00 kcal equals \(8.00 \times 4184\) J, or 33.472 kJ.
03

Applying First Law of Thermodynamics

The First Law of Thermodynamics is given by \(\Delta Q = \Delta U + \Delta W\), where \(\Delta Q\) is the heat added to the system, \(\Delta U\) is the change in internal energy, and \(\Delta W\) is the work done by the system. We need to find \(\Delta U\), the change in internal energy.
04

Formulating the Equation

Substitute the values into the equation: \(\Delta U = \Delta Q - \Delta W\). This becomes \(\Delta U = 33.472\ \text{kJ} - 6.00\ \text{kJ}\).
05

Finding the Change in Internal Energy

Calculate the change in internal energy: \(\Delta U = 33.472 - 6.00 = 27.472\ \text{kJ}\).

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Key Concepts

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

Understanding Internal Energy Change
The concept of internal energy change is pivotal in thermodynamics. Internal energy, denoted as \( \Delta U \), represents the total energy contained within a system. It encompasses all forms of energy present, such as kinetic and potential energy of atoms and molecules. For an isolated system, any change in internal energy is zero; however, an open or closed system can have changes, influenced by heat exchange and work. The formula used to calculate the change in internal energy, derived from the First Law of Thermodynamics, is:
  • \( \Delta U = \Delta Q - \Delta W \)
where \( \Delta Q \) is the heat supplied to the system and \( \Delta W \) is the work done by the system. In our exercise, the calculation \( 33.472\ \text{kJ} - 6.00\ \text{kJ} = 27.472\ \text{kJ} \) shows the increase in internal energy, signifying energy storage within the system.
The Process of Heat Conversion
Heat conversion is crucial when dealing with different units of energy. Here, heat was originally given in kilocalories (kcal), while the standard thermodynamic calculations often require energy to be expressed in joules (J) or kilojoules (kJ). To convert kilocalories to kilojoules:
  • Use the conversion factor: \( 1\ \text{kcal} = 4184\ \text{J} = 4.184\ \text{kJ} \)
Given that the system receives 8.00 kcal, the conversion becomes:
  • \( 8.00\ \text{kcal} \times 4.184\ \text{kJ/kcal} = 33.472\ \text{kJ} \)
This ensures that we use consistent units throughout our calculations. Proper unit conversion is essential to maintain accuracy and coherence in thermodynamic problems.
Understanding Work Done by the System
Work done by a system, expressed as \( \Delta W \), is another significant part of understanding energy changes. When a system performs work, it essentially loses some of its energy to the surroundings. In our scenario, the system does 6.00 kJ of work. For clarity, positive work implies energy is leaving the system — it is energy done by the system to the surroundings, such as expansion of gas against a piston. Therefore, the work done in our problem contributes negatively to the internal energy change, as depicted in the First Law equation:
  • \( \Delta U = \Delta Q - \Delta W \)
This equation accounts for both heat supplied and work done, ensuring we compute how the internal energy adjusts.
Why Energy Units Conversion Matters
Energy conversion, especially converting between kilocalories and kilojoules, is vital due to the differing units encountered in various scientific contexts. The primary reason for conversion is to adhere to the unit consistency required: using either joules or kilojoules in calculations.Kilocalories, often used in biological and nutritional contexts, must be converted to joules or kilojoules for physical calculations. This conversion helps avoid calculation errors and improves understanding. To convert energy units:
  • Remember that \( 1\ \text{kcal} = 4184\ \text{J} = 4.184\ \text{kJ} \)
Using these conversions, we converted 8.00 kcal to 33.472 kJ, which was essential for integrating this value into the broader thermodynamic calculations. Consistent units allow for accurate application of the First Law of Thermodynamics and accurate calculation of energy changes.

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Most popular questions from this chapter

What is the net work output per cycle for the thermodynamic cycle in Fig. \(20-4\) ? We know that the net work output per cycle is the area enclosed by the \(P-V\) diagram. We estimate that in area \(A B C A\) there are 22 squares, each of area $$\left(0.5 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.1 \mathrm{~m}^{3}\right)=5 \mathrm{~kJ}$$ Therefore, Area enclosed by cycle \(\approx(22)(5 \mathrm{~kJ})=1 \times 10^{2} \mathrm{~kJ}\) The net work output per cycle is \(1 \times 10^{2} \mathrm{~kJ}\).

Twenty cubic centimeters of monatomic gas at \(12{ }^{\circ} \mathrm{C}\) and 100 \(\mathrm{kPa}\) is suddenly (and adiabatically) compressed to \(0.50 \mathrm{~cm}^{3}\). Assume that we are dealing with an ideal gas. What are its new pressure and temperature? For an adiabatic change involving an ideal gas, \(P_{1} V_{1}^{\gamma}=P_{2} V_{2}^{\gamma}\) where \(\mathrm{Y}=1.67\) for a monatomic gas. Hence, $$P_{2}=P_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma}=\left(1.00 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(\frac{20}{0.50}\right)^{1.67}=4.74 \times 10^{7} \mathrm{~N} / \mathrm{m}^{2}=47 \mathrm{MPa}$$ To find the final temperature, we could use \(P_{1} V_{1} / T_{1}=P_{2} V_{2} / T_{2}\). Instead, let us use $$T_{1} V_{1}^{\gamma-1}=T_{2} V_{2}^{\gamma}$$ or $$T_{2}=T_{1}\left(\frac{V_{1}}{V_{2}}\right)^{\gamma-1}=(285 \mathrm{~K})\left(\frac{20}{0.50}\right)^{0.67}=(285 \mathrm{~K})(11.8)=3.4 \times 10^{3} \mathrm{~K}$$ As a check, $$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$$ $$ \begin{aligned} \frac{\left(1 \times 10^{5} \mathrm{~N} / \mathrm{m}^{2}\right)\left(20 \mathrm{~cm}^{3}\right)}{285 \mathrm{~K}} &=\frac{\left(4.74 \times 107 \mathrm{~N} / \mathrm{m}^{2}\right)\left(0.50 \mathrm{~cm}^{3}\right)}{3370 \mathrm{~K}} \\ 7000 &=7000 \end{aligned} $$

How much external work is done by an ideal gas in expanding from a volume of \(3.0\) liters to a volume of \(30.0\) liters against a constant pressure of \(2.0\) atm?

A 70 -g metal block moving at \(200 \mathrm{~cm} / \mathrm{s}\) slides across a tabletop a distance of \(83 \mathrm{~cm}\) before it comes to rest. Assuming 75 percent of the thermal energy developed by friction goes into the block, how much does the temperature of the block rise? For the metal, \(c=\) \(0.106 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\)

Find \(\Delta W\) and \(\Delta U\) for a \(6.0\) -cm cube of iron as it is heated from 20 \({ }^{\circ} \mathrm{C}\) to \(300{ }^{\circ} \mathrm{C}\) at atmospheric pressure. For iron, \(c=0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\) and the volume coefficient of thermal expansion is \(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}\) \- 1 . The mass of the cube is 1700 g. Given that \(\Delta T=300^{\circ} \mathrm{C}-20^{\circ} \mathrm{C}=280^{\circ} \mathrm{C}\), \(\Delta Q=c m \Delta T=\left(0.11 \mathrm{cal} / \mathrm{g} \cdot{ }^{\circ} \mathrm{C}\right)(1700 \mathrm{~g})\left(280^{\circ} \mathrm{C}\right)=52 \mathrm{kcal}\) To find that the work done by the expansion of the cube, we need to determine \(\Delta V\). The volume of the cube is \(V=(6.0 \mathrm{~cm})^{3}=216 \mathrm{~cm}^{3} .\) Using \((\Delta V) / V\) \(=\beta \Delta T\), $$\begin{array}{l} \Delta V=V \beta \Delta T=\left(216 \times 10^{-6} \mathrm{~m}^{3}\right)\left(3.6 \times 10^{-5}{ }^{\circ} \mathrm{C}^{-1}\right)\left(280^{\circ} \mathrm{C}\right)=2.18 \times 10^{-6} \mathrm{~m}^{3} \end{array}$$ But the First Law tells us that $$\begin{aligned} \Delta U &=\Delta Q-\Delta W=(52000 \mathrm{cal})(4.184 \mathrm{~J} / \mathrm{cal})-0.22 \mathrm{~J} \\ &=218000 \mathrm{~J}-0.22 \mathrm{~J} \approx 2.2 \times 10^{5} \mathrm{~J} \end{aligned}$$ Notice how very small the work of expansion against the atmosphere is in comparison to \(\Delta U\) and \(\Delta Q .\) Often \(\Delta W\) can be neglected when dealing with liquids and solids.

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