\(\Delta U=100 \mathrm{J}\) for a system that gives off \(100 \mathrm{J}\) of heat and (a) does no work; (b) does 200 J of work; (c) has 100 J of work done on it; (d) has 200 J of work done on it.

Short Answer

Expert verified
The system's internal energy changes as follows: In part (a), \(\Delta U = 100J\); in part (b), \(\Delta U = -100J\); in part (c), \(\Delta U = 200J\); and finally, in part (d), \(\Delta U = 300J\).

Step by step solution

01

Part (a) Calculations

For part (a), the system does no work. This means that \(W = 0\). Plugging this into the formula, we get \(\Delta U = Q - W = 100J - 0 = 100J\)
02

Part (b) Calculations

For part (b), the system does 200J of work. So, \(W = 200J\). Substitute this into the formula to get \(\Delta U = Q - W = 100J - 200J = -100J\)
03

Part (c) Calculations

For part (c), the system has 100J of work done on it. In this case, the work done on the system is negative, hence \(W = -100J\). Substituting this into the equation, we find \(\Delta U = Q - W = 100J - -100J = 200J\)
04

Part (d) Calculations

In part (d), the system has 200J of work done on it. Here, the work done on the system is also negative, so \(W = -200J\). This gives \(\Delta U = Q - W = 100J - -200J = 300J\)

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Key Concepts

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

Internal Energy Change
The concept of Internal Energy Change is fundamental in understanding thermodynamics. It represents the total change in a system's energy state. This change, denoted by \( \Delta U \), can result from the energy the system exchanges with its surroundings. It can occur in the form of heat \( Q \) or work \( W \) done on or by the system.

In our exercise, for instance, the system is said to have a change in internal energy of \( \Delta U = 100 \, \text{J} \). This change is resultant of the heat and work interactions with the environment. When no work is done by or on the system, it simply means all of the energy change is due to heat transfer, leaving the internal energy increased by the heat amount \( Q \).
Work Done by System
When discussing the topic of Work Done by System, it's essentially energy transferred by the system to its surroundings. If the system expands against an external pressure, for example, it does work on the surroundings, which leads to a decrease in the system's internal energy.

In the exercise, when the system does 200J of work, it loses that amount of energy, hence the negative sign in the mathematical expression \( W = 200J \). This action decreases its internal energy, which is why we calculate \( \Delta U \) as being \( -100J \) in part (b). The key is to recognize that work done by the system reduces its internal energy.
Heat Transfer
Heat Transfer is energy in transit due to a temperature difference. Whenever heat is added to a system, its internal energy increases unless the system does work to dissipate the energy. Conversely, if the system loses heat, its internal energy decreases.

In the context of our exercise, the system releases \( 100 \, \text{J} \) of heat. If no work is done, as in part (a), that energy is directly subtracted from the system's internal energy. However, if work is involved, as in parts (b), (c), and (d), we must consider both heat and work to determine the net change in internal energy.
Thermodynamic Processes
Exploring Thermodynamic Processes reveals how systems evolve between different states. These processes involve changes in internal energy (\( \Delta U \)), heat transfer (\( Q \)), and the work done (\( W \)) by or on the system. The first law of thermodynamics is the guiding principle, stating that energy can neither be created nor destroyed, only transformed.

The processes described in our exercise represent different thermodynamic scenarios. In parts (a) and (b), the system evolves without external work, while in (c) and (d), it experiences work done on it, illustrated by the positive change in internal energy. These variations show how energy balance is maintained according to the first law of thermodynamics.

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Most popular questions from this chapter

In the Are You Wondering \(7-1\) box, the temperature variation of enthalpy is discussed, and the equation \(q_{P}=\) heat capacity \(\times\) temperature change \(=C_{P} \times \Delta T\) was introduced to show how enthalpy changes with temperature for a constant-pressure process. Strictly speaking, the heat capacity of a substance at constant pressure is the slope of the line representing the variation of enthalpy (H) with temperature, that is $$C_{P}=\frac{d H}{d T} \quad(\text { at constant pressure })$$ where \(C_{P}\) is the heat capacity of the substance in question. Heat capacity is an extensive quantity and heat capacities are usually quoted as molar heat capacities \(C_{P, \mathrm{m}},\) the heat capacity of one mole of substance; an intensive property. The heat capacity at constant pressure is used to estimate the change in enthalpy due to a change in temperature. For infinitesimal changes in temperature, $$d H=C_{p} d T \quad(\text { at constant pressure })$$ To evaluate the change in enthalpy for a particular temperature change, from \(T_{1}\) to \(T_{2}\), we write $$\int_{H\left(T_{1}\right)}^{H\left(T_{2}\right)} d H=H\left(T_{2}\right)-H\left(T_{1}\right)=\int_{T_{1}}^{T_{2}} C_{P} d T$$ If we assume that \(C_{P}\) is independent of temperature, then we recover equation (7.5) $$\Delta H=C_{P} \times \Delta T$$ On the other hand, we often find that the heat capacity is a function of temperature; a convenient empirical expression is $$C_{P, \mathrm{m}}=a+b T+\frac{c}{T^{2}}$$ What is the change in molar enthalpy of \(\mathrm{N}_{2}\) when it is heated from \(25.0^{\circ} \mathrm{C}\) to \(100.0^{\circ} \mathrm{C} ?\) The molar heat capacity of nitrogen is given by$$C_{P, \mathrm{m}}=28.58+3.77 \times 10^{-3} T-\frac{0.5 \times 10^{5}}{T^{2}} \mathrm{JK}^{-1} \mathrm{mol}^{-1}$$

What is the change in internal energy of a system if the system (a) absorbs \(58 \mathrm{J}\) of heat and does \(58 \mathrm{J}\) of work; (b) absorbs 125 J of heat and does 687 J of work; (c) evolves 280 cal of heat and has 1.25 kJ of work done on it?

A calorimeter that measures an exothermic heat of reaction by the quantity of ice that can be melted is called an ice calorimeter. Now consider that \(0.100 \mathrm{L}\) of methane gas, \(\mathrm{CH}_{4}(\mathrm{g}),\) at \(25.0^{\circ} \mathrm{C}\) and \(744 \mathrm{mm} \mathrm{Hg}\) is burned at constant pressure in air. The heat liberated is captured and used to melt \(9.53 \mathrm{g}\) ice at \(0^{\circ} \mathrm{C}\left(\Delta H_{\text {fusion }} \text { of ice }=6.01 \mathrm{kJ} / \mathrm{mol}\right)\) (a) Write an equation for the complete combustion of \(\mathrm{CH}_{4},\) and show that combustion is incomplete in this case. (b) Assume that \(\mathrm{CO}(\mathrm{g})\) is produced in the incomplete combustion of \(\mathrm{CH}_{4}\), and represent the combustion as best you can through a single equation with small whole numbers as coefficients. \((\mathrm{H}_{2} \mathrm{O}(\mathrm{l})\) is another . product of the combustion.)

Calculate the final temperature that results when (a) a 12.6 g sample of water at \(22.9^{\circ} \mathrm{C}\) absorbs \(875 \mathrm{J}\) of heat; (b) a 1.59 kg sample of platinum at \(78.2^{\circ} \mathrm{C}\) gives off \(1.05 \mathrm{kcal}\) of heat \(\left(\mathrm{sp} \mathrm{ht} \text { of } \mathrm{Pt}=0.032 \mathrm{cal} \mathrm{g}^{-1}\right.\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\).

You are planning a lecture demonstration to illustrate an endothermic process. You want to lower the temperature of \(1400 \mathrm{mL}\) water in an insulated container from 25 to \(10^{\circ} \mathrm{C} .\) Approximately what mass of \(\mathrm{NH}_{4} \mathrm{Cl}(\mathrm{s})\) should you dissolve in the water to achieve this result? The heat of solution of \(\mathrm{NH}_{4} \mathrm{Cl}\) is \(+14.7 \mathrm{kJ} / \mathrm{mol} \mathrm{NH}_{4} \mathrm{Cl}\).

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