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Chapter 6: Problem 31

If the internal energy of a thermodynamic system is increased by \(300 .\) 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
The heat transferred to the thermodynamic system is \(375\,\text{J}\).

Step by step solution

01

Write down the equation for the first law of thermodynamics.

The first law of thermodynamics equation can be written as: \[ \Delta U = Q - W \] Where: - \(\Delta U\) is the change in internal energy. - \(Q\) is the amount of heat transferred. - \(W\) is the work done.
02

Substitute the given values into the equation.

We are given the values for the internal energy increase (\(\Delta U = 300\,\text{J}\)) and the expansion work done (\(W = 75\,\text{J}\)). Plugging these values into the equation, we get: \[ 300\,\text{J} = Q - 75\,\text{J} \]
03

Solve for the heat transferred.

To find the amount of heat transferred (\(Q\)), we need to isolate \(Q\) in our equation. We do this by adding \(75\,\text{J}\) to both sides of the equation: \[ 300\,\text{J} + 75\,\text{J} = Q \] Now, we simply add the values on the left side: \[ 375\,\text{J} = Q \]
04

Determine the direction of the heat transfer.

Since the value of \(Q\) is positive (\(375\,\text{J}\)), it means that the heat was transferred to the system. If the value of \(Q\) were negative, then it would have meant that the heat was transferred from the system. So, the heat transferred to the thermodynamic system is \(375\,\text{J}\).

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Key Concepts

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

Internal Energy
Understanding internal energy is crucial in thermodynamics. It refers to the total energy stored within a system. This energy is a sum of kinetic and potential energies of all particles in a system. Internal energy is a state function, meaning it depends only on the current state of the system, not how that state was reached.

In this exercise, the internal energy increased by 300 J. When we talk about an increase in internal energy, it means that the system gained additional energy. This can happen through processes like heat transfer or work being done on the system. A key point to remember is that internal energy changes with conditions such as temperature, pressure, and volume, as well as through chemical reactions within the system.
Expansion Work
Expansion work is a term in thermodynamics that refers to the work done when a system changes its volume under pressure. Simply put, when a gas expands, it pushes against external pressure, and this effort is called expansion work. This work can also be quantified as the product of pressure and volume change, commonly given by the formula: \[ W = P imes \Delta V \] where \(P\) stands for pressure and \(\Delta V\) represents the change in volume.

In the given problem, 75 J of expansion work is performed by the system. It indicates that the system used 75 J of energy to accomplish this expansion. This energy expenditure has to be considered when calculating the overall energy changes in the system and plays a crucial role in understanding how energy is conserved as it transitions between different forms.
Heat Transfer
Heat transfer is one of the primary ways energy is exchanged between a system and its surroundings. Within thermodynamics, it is crucial to understand whether energy is entering or leaving the system. The sign of the heat transfer value tells us this 鈥� positive for heat entering the system and negative for heat exiting.

In the exercise, calculating the heat transfer is about applying the first law of thermodynamics, which is represented as:\[ \Delta U = Q - W \] Using this formula, we found the heat transfer (\(Q\)) to be 375 J. Since it is a positive quantity, it indicates that the environment supplied 375 J of heat to the system. Recognizing the direction of heat transfer helps in comprehending how systems reach equilibrium and react to changes in their surroundings, which is pivotal in scenarios ranging from engine performance to climate models.

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Most popular questions from this chapter

Combustion of table sugar produces \(\mathrm{CO}_{2}(g)\) and \(\mathrm{H}_{2} \mathrm{O}(l) .\) When \(1.46 \mathrm{~g}\) table sugar is combusted in a constant-volume (bomb) calorimeter, \(24.00 \mathrm{~kJ}\) of heat is liberated. a. Assuming that table sugar is pure sucrose, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)\), write the balanced equation for the combustion reaction. b. Calculate \(\Delta E\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose. c. Calculate \(\Delta I I\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) for the combustion reaction of sucrose at \(25^{\circ} \mathrm{C}\).

A system releases \(125 \mathrm{~kJ}\) of heat while \(104 \mathrm{~kJ}\) of work is done on it. Calculate \(\Delta E\).

You have a \(1.00\) -mol sample of water at \(-30 .{ }^{\circ} \mathrm{C}\) and you heat it until you have gaseous water at \(140 .^{\circ} \mathrm{C}\). Calculate \(q\) for the entire process. Use the following data. Specific heat capacity of ice \(=2.03 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of water \(=4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) Specific heat capacity of steam \(=2.02 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) \(\begin{array}{lr}\mathrm{H}_{2} \mathrm{O}(s) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) & \Delta H_{\text {fusion }}=6.02 \mathrm{~kJ} / \mathrm{mol}\left(\text { at } 0^{\circ} \mathrm{C}\right) \\\ \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{2} \mathrm{O}(g) & \Delta H_{\text {vaporization }}=40.7 \mathrm{~kJ} / \mathrm{mol}\left(\text { at } 100 .{ }^{\circ} \mathrm{C}\right)\end{array}\)

Nitromethane, \(\mathrm{CH}_{3} \mathrm{NO}_{2}\), can be used as a fuel. When the liquid is burned, the (unbalanced) reaction is mainly $$ \mathrm{CH}_{3} \mathrm{NO}_{2}(l)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g)+\mathrm{N}_{2}(g)+\mathrm{H}_{2} \mathrm{O}(g) $$ a. The standard enthalpy change of reaction \(\left(\Delta H_{\mathrm{rxn}}^{\circ}\right)\) for the balanced reaction (with lowest whole- number coefficients) is \(-1288.5 \mathrm{~kJ} .\) Calculate the \(\Delta H_{\mathrm{f}}^{\circ}\) for nitromethane. b. A \(15.0\) - \(\mathrm{L}\) flask containing a sample of nitromethane is filled with \(\mathrm{O}_{2}\) and the flask is heated to \(100 .^{\circ} \mathrm{C}\). At this temperature, and after the reaction is complete, the total pressure of all the gases inside the flask is 950 . torr. If the mole fraction of nitrogen ( \(\chi_{\text {nitrogen }}\) ) is \(0.134\) after the reaction is complete, what mass of nitrogen was produced?

A cubic piece of uranium metal (specific heat capacity \(=0.117\) \(\mathrm{J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) at \(200.0^{\circ} \mathrm{C}\) is dropped into \(1.00 \mathrm{~L}\) deuterium oxide ("heavy water," specific heat capacity \(=4.211 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\) ) at \(25.5^{\circ} \mathrm{C}\). The final temperature of the uranium and deuterium oxide mixture is \(28.5^{\circ} \mathrm{C}\). Given the densities of uranium \(\left(19.05 \mathrm{~g} / \mathrm{cm}^{3}\right)\) and deuterium oxide (1.11 \(\mathrm{g} / \mathrm{mL}\) ), what is the edge length of the cube of uranium?
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