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Chapter 6: Problem 37

If the internal energy of a thermodynamic system is increased by \(300 . \mathrm{J}\) while 75 \(\mathrm{J}\) of expansion work is done, how much heat was transferred and in which direction, to or from the system?

Short Answer

Expert verified
In this thermodynamic process, 375 J of heat was transferred to the system.

Step by step solution

01

Identify the known and unknown variables

We are given: 1. The increase in internal energy, 鈭哢 = 300 J 2. Work done by the system, W = 75 J We need to find the heat transferred, Q, and its direction (to or from the system).
02

Write down the equation for the first law of thermodynamics

The first law of thermodynamics for a closed system can be stated as: 鈭哢 = Q - W where 鈭哢 is the change in internal energy, Q is the heat transferred, and W is the work done by the system.
03

Substitute the known values and find Q

Now we can substitute the given values into the equation: \(300 \text{ J} = Q - 75 \text{ J}\) To solve for Q, add 75 J to both sides of the equation: \(Q = 300 \text{ J} + 75 \text{ J}\)
04

Calculate the heat transferred and determine its direction

Do the arithmetic to find the value of Q: \(Q = 375 \text{ J}\) The sign of Q is positive, which indicates that heat was transferred to the system. If Q was negative, that would mean heat was transferred from the system.
05

Final Answer

In this thermodynamic process, 375 J of heat was transferred to the system.

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Key Concepts

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

Understanding Internal Energy
Internal energy is a fundamental concept in thermodynamics that represents the total energy contained within a system. It consists of microscopic components such as the kinetic and potential energies of atoms and molecules.
When we talk about a system's internal energy change, we denote it as \( \Delta U \). This change can occur due to heat added to or removed from the system, or work done on or by the system.
  • An increase in internal energy usually means that the system absorbed energy, like in our exercise where \( \Delta U = 300 \, \mathrm{J} \).
  • A decrease means that the system released energy.
Internal energy is crucial because it sums up various energy forms in a system, helping in analyzing processes under the first law of thermodynamics.
The Role of Heat Transfer
Heat transfer refers to the movement of thermal energy from one system or body to another due to a temperature difference. It's one of the key ways energy is exchanged in thermodynamics.
Heat transferred is denoted by \( Q \). In our exercise, we needed to find \( Q \) to determine how energy shifted in and out of the system.
Here's a quick briefing on its direction:
  • If \( Q \) is positive, heat has been transferred to the system.
  • If \( Q \) is negative, it has been transferred from the system.
In the context of our problem, we found \( Q = 375 \, \mathrm{J} \), indicating that heat was transferred into the system, further increasing its internal energy.
Work Done on or by the System
Work done is another pivotal component of energy change in thermodynamics. In layman's terms, it's energy exerted when a force causes displacement. In thermodynamic processes, this often relates to expansion or compression work.
Work done (\( W \)) can be positive or negative:
  • When the system does work on its surroundings, \( W \) is considered positive, like the 75 J in our example.
  • If work is done on the system, \( W \) is negative.
Knowing whether the work done is positive or negative helps determine how energy transfers influence a system, as well as understanding energy conservation under the first law of thermodynamics. For our step-by-step problem, understanding work done helped deduce the heat (\( Q \)) transferred within the system.

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Most popular questions from this chapter

One of the components of polluted air is NO. It is formed in the high- temperature environment of internal combustion engines by the following reaction: $$ \mathrm{N}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{NO}(g) \quad \Delta H=180 \mathrm{kJ} $$ Why are high temperatures needed to convert \(\mathrm{N}_{2}\) and \(\mathrm{O}_{2}\) to NO?

A serving size of six cookies contains 4 g of fat, 20 of carbohydrates, and 2 g of protein. If walking 1.0 mile consumes 170 kJ of energy, how many miles must you walk to burn off enough calories to eat six cookies? Assume the energy content of fats, carbohydrates, and proteins are 8 kcallg, 4 kcallg, and 4 kcallg, respectively.

Consider 2.00 moles of an ideal gas that are taken from state \(A\) \(\left(P_{A}=2.00 \mathrm{atm}, V_{A}=10.0 \mathrm{L}\right)\) to state \(B\left(P_{B}=1.00 \mathrm{atm}, V_{B}=\right.\) 30.0 \(\mathrm{L}\) ) by two different pathways: These pathways are summarized on the following graph of \(P\) versus \(V :\) Calculate the work (in units of \(\mathrm{J} )\) associated with the two path- ways. Is work a state function? Explain.

A gaseous hydrocarbon reacts completely with oxygen gas to form carbon dioxide and water vapor. Given the following data, determine \(\Delta H_{f}^{\circ}\) for the hydrocarbon: $$ \begin{aligned} \Delta H_{\mathrm{reacion}}^{\circ} &=-2044.5 \mathrm{kJ} / \mathrm{mol} \text { hydrocarbon } \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{CO}_{2}\right) &=-393.5 \mathrm{kJ} / \mathrm{mol} \\ \Delta H_{\mathrm{f}}^{\circ}\left(\mathrm{H}_{2} \mathrm{O}\right) &=-242 \mathrm{kJ} / \mathrm{mol} \end{aligned} $$ Density of \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) product mixture at 1.00 \(\mathrm{atm}\) , \(200 . \mathrm{C}=0.751 \mathrm{g} / \mathrm{L}\) . The density of the hydrocarbon is less than the density of Kr at the same conditions.

Consider the reaction $$ 2 \mathrm{HCl}(a q)+\mathrm{Ba}(\mathrm{OH})_{2}(a q) \longrightarrow \mathrm{BaCl}_{2}(a q)+2 \mathrm{H}_{2} \mathrm{O}(l) $$ $$ \Delta H=-118 \mathrm{kJ} $$ Calculate the heat when 100.0 \(\mathrm{mL}\) of 0.500\(M \mathrm{HCl}\) is mixed with 300.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{Ba}(\mathrm{OH})_{2}\) . Assuming that the temperature of both solutions was initially \(25.0^{\circ} \mathrm{C}\) and that the final mixture has a mass of 400.0 \(\mathrm{g}\) and a specific heat capacity of 4.18 \(\mathrm{J} / \mathrm{C} \cdot \mathrm{g}\) , calculate the final temperature of the mixture.
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