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Calculate \(\Delta E\) for the following situations: a. \(q=120.0 \mathrm{J} ; w=-40.0 \mathrm{J}\) b. \(q=9.2 \mathrm{kJ} ; w=0.70 \mathrm{J}\) c. \(q=-625 \mathrm{J} ; w=-315 \mathrm{J}\)

Short Answer

Expert verified
Question: Calculate the change in internal energy (\(\Delta E\)) for each of the following situations: a) \(q=120.0 \mathrm{J}\) and \(w=-40.0 \mathrm{J}\) b) \(q=9.2 \, \mathrm{kJ}\) and \(w=0.70 \, \mathrm{J}\) c) \(q=-625 \, \mathrm{J}\) and \(w=-315 \, \mathrm{J}\) Answer: a) \(\Delta E = 80.0 \mathrm{J}\) b) \(\Delta E = 9200.7 \mathrm{J}\) c) \(\Delta E = -940 \mathrm{J}\)

Step by step solution

01

Part a: Calculating \(\Delta E\) for the first situation

We are given \(q=120.0 \mathrm{J}\) and \(w=-40.0 \mathrm{J}\). Using the formula above, we have: $$\Delta E = q + w = 120.0 \mathrm{J} - 40.0 \mathrm{J} = 80.0 \mathrm{J}.$$
02

Part b: Calculating \(\Delta E\) for the second situation

We are given \(q=9.2 \, \mathrm{kJ} (= 9200 \, \mathrm{J})\) and \(w=0.70 \, \mathrm{J}\). Using the formula above, we have: $$\Delta E = q + w = 9200 \mathrm{J} + 0.70 \mathrm{J} = 9200.7 \mathrm{J}.$$
03

Part c: Calculating \(\Delta E\) for the third situation

We are given \(q=-625 \, \mathrm{J}\) and \(w=-315 \, \mathrm{J}\). Using the formula above, we have: $$\Delta E = q + w = -625 \mathrm{J} - 315 \mathrm{J} = -940 \mathrm{J}.$$

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Key Concepts

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

Energy Change Calculation
In thermodynamics, calculating the change in energy, denoted by \( \Delta E \), is essential to understanding how energy is transformed in different systems. When examining changes in energy, we consider two critical factors: heat \((q)\) and work \((w)\). Both of these elements contribute to the total change in energy of a system. The formula we use is:\[ \Delta E = q + w \]This formula tells us that the change in energy is the sum of the heat added to or removed from the system and the work done on or by the system. Here are some important points to remember:
  • When \( q > 0 \), heat is added to the system, thus, increasing the energy.
  • When \( q < 0 \), heat is removed from the system, causing a decrease in energy.
  • For work: \( w > 0 \), work is done on the system, and \( w < 0 \), work is done by the system.
In the exercise, you see how simply adding the given values of \( q \) and \( w \) provides the energy change for different scenarios. Each calculation uses the straightforward concept that energy change equals heat plus work.
First Law of Thermodynamics
The First Law of Thermodynamics is a fundamental principle in chemistry and physics that states energy cannot be created or destroyed, only transformed. This is essentially a statement of the conservation of energy. The first law is expressed in the equation: \[ \Delta E = q + w \]This principle helps us understand how energy flows and converts between systems and surroundings. For any closed system, any energy loss by the system must equal energy gain by the surroundings, and vice versa. Key points include:
  • If a system does work on its surroundings, it loses energy, hence \( w < 0 \).
  • Conversely, if the surroundings do work on the system, the system gains energy, causing \( w > 0 \).
Understanding the First Law helps predict how energy will behave in responses discussed in scenarios like chemical reactions, where heat exchange and work interaction are fundamental.
Work and Heat in Chemistry
In chemistry, work and heat are two forms of energy transfer and are crucial for understanding thermodynamic processes. Let's explore how they function:**Heat** - Heat is the transfer of thermal energy between systems or surroundings because of a temperature difference.- Heat flows spontaneously from the hot region to the cold one.- In chemical reactions, heat can be released or absorbed, affecting the internal energy and possibly the reaction rate.**Work** - In the context of chemistry, work often involves mechanical or electrical energy transfer.- In our exercise, work is particularly important when considering gas expansions or compressions, where work against external pressures alters the internal energy.Both work and heat contribute to changes in a system's total energy. For calculations, notice how they directly impact the net change in energy, as they are the sole components in the energy change equation \( \Delta E = q + w \). By mastering these concepts, one can delve deeper into the energetics and dynamics of chemical processes.

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Most popular questions from this chapter

Use the following information to calculate the enthalpy change involved in the complete reaction of \(3.0 \mathrm{g}\) of carbon to form \(\mathrm{PbCO}_{3}(s)\) in reaction \(4 .\) Be sure to give the proper sign (positive or negative) with your answer. (1) \(\quad \mathrm{Pb}(s)+\frac{1}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{PbO}(s) \quad \Delta H_{\mathrm{rm}}^{\circ}=-219 \mathrm{kJ}\) (2) \(\quad \mathrm{C}(s)+\mathrm{O}_{2}(g) \rightarrow \mathrm{CO}_{2}(g) \quad \quad \Delta H_{\mathrm{rxn}}^{\circ}=-394 \mathrm{kJ}\) (3) \(\quad \mathrm{PbCO}_{3}(s) \rightarrow \mathrm{PbO}(s)+\mathrm{CO}_{2}(g) \quad \Delta H_{\mathrm{rxn}}^{\circ}=86 \mathrm{kJ}\) (4) \(\quad \mathrm{Pb}(s)+\mathrm{C}(s)+\frac{3}{2} \mathrm{O}_{2}(g) \rightarrow \mathrm{PbCO}_{3}(s) \quad \Delta H_{\mathrm{rsn}}^{\circ}=?\)

Ammonium nitrate decomposes to \(\mathrm{N}_{2} \mathrm{O}\) and water vapor at temperatures between \(250^{\circ} \mathrm{C}\) and \(300^{\circ} \mathrm{C} .\) Write a balanced chemical reaction describing the decomposition of ammonium nitrate, and calculate the enthalpy of reaction by using the appropriate enthalpies of formation from Appendix 4.

The specific heat of solid copper is \(0.385 \mathrm{J} / \mathrm{g} \cdot^{\circ} \mathrm{C}\) ). What thermal energy change occurs when a \(35.3 \mathrm{g}\) sample of copper is cooled from \(35.0^{\circ} \mathrm{C}\) to \(15.0^{\circ} \mathrm{C} ?\) Be sure to give your answer the proper sign. This amount of energy is used to melt solid ice at \(0.0^{\circ} \mathrm{C} .\) The molar enthalpy of fusion of ice is \(6.01 \mathrm{kJ} / \mathrm{mol} .\) How many moles of ice are melted?

Automobile air bags produce nitrogen gas from the reaction: $$2 \mathrm{NaN}_{3}(s) \rightarrow 2 \mathrm{Na}(s)+3 \mathrm{N}_{2}(g)$$ a. If \(2.25 \mathrm{g}\) of \(\mathrm{NaN}_{3}\) reacts to fill an air bag, how much \(P-V\) work will the \(\mathrm{N}_{2}\) do against an external pressure of 1.00 atm given that the density of nitrogen is \(1.165 \mathrm{g} / \mathrm{L}\) at \(20^{\circ} \mathrm{C} ?\) b. If the process releases \(2.34 \mathrm{kJ}\) of heat, what is \(\Delta E\) for the system?

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