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What is the change in internal energy of a system if the surroundings (a) transfer 235 J of heat and 128 J of work to the system; (b) absorb 145 J of heat from the system while doing \(98 \mathrm{J}\) of work on the system; (c) exchange no heat, but receive 1.07 kJ of work from the system?

Short Answer

Expert verified
The change in internal energy for the system is (a) 107 J, (b) -243 J, (c) 1070 J

Step by step solution

01

Calculate the change in internal energy for part (a)

For part (a), both heat (Q=235 J) and work (W=128 J) are being transferred to the system. We simply replace those values into the formula: \(ΔU = Q - W\) so it results in \(ΔU = 235 - 128 = 107 J\).
02

Calculate the change in internal energy for part (b)

For part (b), the surroundings are absorbing heat from the system (Q=-145 J) and doing work on it (W=98 J). Plug these values into the equation. However, be sure to make the heat value negative, as it is being taken away from the system: \(ΔU = -145 -98 = -243 J\).
03

Calculate the change in internal energy for part (c)

For part (c), there is no exchange of heat (Q=0) and the system is doing work (W=-1.07 kJ or -1070 J as we need to convert kJ into J). Plugging these values into our formula we get: \(ΔU = 0 - (-1070) = 1070 J\).

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Key Concepts

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

First Law of Thermodynamics
In the realm of thermodynamics, understanding how energy flows within a system and its surroundings is governed by the First Law of Thermodynamics. It's a fundamental principle stating that energy cannot be created or destroyed, but merely transformed from one form to another.
In practical terms, this means that the change in internal energy of a system ( ΔU = Q - W ) can be calculated by accounting for the heat added to the system ( Q ) and the work done by the system ( W ).
In our original exercise, this principle provides clarity: simply plug in the values of heat and work to determine how the system's internal energy is altered.
Heat Transfer
Heat transfer is the process by which thermal energy moves from one object or system to another. This can occur in several forms such as conduction, convection, and radiation.
In the context of thermodynamics, we focus on how much heat energy ( Q ) is added to or removed from a system.
- If heat energy is added (positive Q ), the internal energy typically increases. - If heat energy is removed (negative Q ), the internal energy decreases.
In part (b) of the original solution, for instance, the system loses heat while part (a) gains it, illustrating how heat transfer directly impacts the total internal energy of a system.
Work in Thermodynamics
Work in thermodynamics refers to the energy transferred when an external force moves a boundary of the system.
This is captured in the formula for internal energy change as work done on or by the system ( W ).
The sign of W can vary:
  • If work is done on the system, W is positive.
  • If work is done by the system, W is negative.
In the exercise's part (c), note that no heat is exchanged but the system performs work on the surroundings, resulting in a negative W value that significantly alters internal energy.

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Most popular questions from this chapter

How much heat, in kilojoules, is associated with the production of \(283 \mathrm{kg}\) of slaked lime, \(\mathrm{Ca}(\mathrm{OH})_{2} ?\) $$\mathrm{CaO}(\mathrm{s})+\mathrm{H}_{2} \mathrm{O}(1) \longrightarrow \mathrm{Ca}(\mathrm{OH})_{2}(\mathrm{s}) \quad \Delta H^{\circ}=-65.2 \mathrm{kJ}$$

An alternative approach to bomb calorimetry is to establish the heat capacity of the calorimeter, exclusive of the water it contains. The heat absorbed by the water and by the rest of the calorimeter must be calculated separately and then added together. A bomb calorimeter assembly containing \(983.5 \mathrm{g}\) water is calibrated by the combustion of \(1.354 \mathrm{g}\) anthracene. The temperature of the calorimeter rises from 24.87 to \(35.63^{\circ} \mathrm{C} .\) When \(1.053 \mathrm{g}\) citric acid is burned in the same assembly, but with 968.6 g water, the temperature increases from 25.01 to \(27.19^{\circ} \mathrm{C}\). The heat of combustion of anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10}(\mathrm{s}),\) is \(-7067 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{C}_{14} \mathrm{H}_{10} \cdot\) What is the heat of combustion of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) expressed in \(\mathrm{kJ} / \mathrm{mol} ?\)

\(\mathrm{CCl}_{4},\) an important commercial solvent, is prepared by the reaction of \(\mathrm{Cl}_{2}(\mathrm{g})\) with a carbon compound. Determine \(\Delta H^{\circ}\) for the reaction $$ \mathrm{CS}_{2}(1)+3 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CCl}_{4}(1)+\mathrm{S}_{2} \mathrm{Cl}_{2}(1) $$ Use appropriate data from the following listing. $$\begin{aligned} \mathrm{CS}_{2}(\mathrm{l})+3 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{SO}_{2}(\mathrm{g}) & \\ \Delta H^{\circ}=&-1077 \mathrm{kJ} \end{aligned}$$ $$2 \mathrm{S}(\mathrm{s})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{S}_{2} \mathrm{Cl}_{2}(1) \quad \Delta H^{\circ}=-58.2 \mathrm{kJ}$$ $$\mathrm{C}(\mathrm{s})+2 \mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{CCl}_{4}(1) \quad \Delta H^{\circ}=-135.4 \mathrm{kJ}$$ $$\mathrm{S}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-296.8 \mathrm{kJ}$$ $$\mathrm{SO}_{2}(\mathrm{g})+\mathrm{Cl}_{2}(\mathrm{g}) \longrightarrow \mathrm{SO}_{2} \mathrm{Cl}_{2}(1) \quad \Delta H^{\circ}=+97.3 \mathrm{kJ}$$ $$\mathrm{C}(\mathrm{s})+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-393.5 \mathrm{kJ}$$ $$\begin{aligned} \mathrm{CCl}_{4}(1)+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{COCl}_{2}(\mathrm{g})+\mathrm{Cl}_{2} \mathrm{O}(\mathrm{g}) & \\ \Delta H^{\circ}=&-5.2 \mathrm{kJ} \end{aligned}$$

The standard molar enthalpy of formation of \(\mathrm{CO}_{2}(\mathrm{g})\) is equal to (a) \(0 ;\) (b) the standard molar heat of combustion of graphite; (c) the sum of the standard molar enthalpies of formation of \(\mathrm{CO}(\mathrm{g})\) and \(\mathrm{O}_{2}(\mathrm{g})\) (d) the standard molar heat of combustion of \(\mathrm{CO}(\mathrm{g})\)

A 1.103 g sample of a gaseous carbon-hydrogenoxygen compound that occupies a volume of \(582 \mathrm{mL}\) at 765.5 Torr and \(25.00^{\circ} \mathrm{C}\) is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb calorimeter. The products of the combustion are \(2.108 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}), 1.294 \mathrm{g} \mathrm{H}_{2} \mathrm{O}(1),\) and enough heat to raise the temperature of the calorimeter assembly from 25.00 to \(31.94^{\circ} \mathrm{C}\). The heat capacity of the calorimeter is \(5.015 \mathrm{kJ} /^{\circ} \mathrm{C}\). Write an equation for the combustion reaction, and indicate \(\Delta H^{\circ}\) for this reaction at \(25.00^{\circ} \mathrm{C}\).

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