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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?

Short Answer

Expert verified
The change in internal energy of the system are (a) 0J, (b) -562J, and (c) 2421.52J.

Step by step solution

01

Calculate Internal Energy Change for Part (a)

Use the first law of Thermodynamics formula which is \( \Delta U = Q - W \), where \( \Delta U \) is the change in internal energy, Q is the heat absorbed by the system and W is the work done by the system. Substituting the given values, we get \( \Delta U = 58J - 58J = 0J \).
02

Calculate Internal Energy Change for Part (b)

Similarly for part (b), using the first law of Thermodynamics formula, \( \Delta U = Q - W \). Substitute the given values, \( \Delta U = 125J - 687J = -562J \). The negative sign indicates that energy has left the system.
03

Convert Calories to Joules for Part (c)

Given 1 cal = 4.184 J. So, 280 cal would be \( 280cal * 4.184 J/cal = 1171.52J \).
04

Calculate Internal Energy Change for Part (c)

Now, compute the change in internal energy for part (c). We need to remember that the work is done on the system this time, so it's considered negative. Converting work done from kJ to J (since 1 kJ = 1000 J), we get \( 1.25kJ * 1000 J/kJ = 1250J \). Hence, interior energy change is calculated as \( \Delta U = Q - W = 1171.52J - (-1250J) = 2421.52J \).

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Key Concepts

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

Internal Energy
When we talk about internal energy, we're referring to the total energy contained within a system. This includes the kinetic energy of molecules moving about and the potential energy associated with them. Internal energy is crucial in understanding how energy transfers affect a system.

One essential point is that the internal energy of a system can change due to exchanges of heat and work. If you ever find yourself wondering why a system feels hotter or colder, or why it seems to move with more or less vigor, it might be because its internal energy has shifted.
  • Increase in internal energy means the system becomes more energetic.
  • Decrease in internal energy generally implies the system loses some energy to its surroundings.
By observing internal energy, we can better predict and describe the behavior of physical and chemical systems under various conditions.
Heat Absorption
Heat absorption is a critical concept when studying energy changes within a system. It is one way energy can be transferred into the system, causing a warming effect or fueling a reaction.

When a system absorbs heat:
  • The temperature of the system may increase, although in some cases, heat might cause changes in state without altering temperature.
  • Potential energy within the system could increase, depending on the system's characteristics.
The sign convention in the first law of thermodynamics states that absorbed heat is considered positive (as it's entering the system). If a system absorbs more heat than it loses, you can expect its internal energy to rise, which might affect the system's overall activity.
Work Done on System
The concept of work done on a system is another path by which energy can influence a system's internal energy. Whenever an external force performs work on a system, it essentially adds energy to it.

In thermodynamics:
  • If work is done on a system, this work is treated as negative in calculations because it adds energy to the system.
  • If the system itself does work on its surroundings, this work is positive as it expends energy from the system.
For a practical example, consider when you compress a gas inside a cylinder: work is done on the gas, which increases its internal energy. Hence, a system that has work done on it tends to experience a rise in its energy level, possibly influencing its temperature, volume, or phase.

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Most popular questions from this chapter

Explain the important distinctions between each pair of terms: (a) system and surroundings; (b) heat and work; (c) specific heat and heat capacity; (d) endothermic and exothermic; (e) constant-volume process and constant-pressure process.

The combustion of methane gas, the principal constituent of natural gas, is represented by the equation $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ \Delta H^{\circ} &=-890.3 \mathrm{kJ} \end{aligned}$$ (a) What mass of methane, in kilograms, must be burned to liberate \(2.80 \times 10^{7} \mathrm{kJ}\) of heat? (b) What quantity of heat, in kilojoules, is liberated in the complete combustion of \(1.65 \times 10^{4} \mathrm{L}\) of \(\mathrm{CH}_{4}(\mathrm{g})\) measured at \(18.6^{\circ} \mathrm{C}\) and \(768 \mathrm{mmHg} ?\) (c) If the quantity of heat calculated in part (b) could be transferred with \(100 \%\) efficiency to water, what volume of water, in liters, could be heated from 8.8 to \(60.0^{\circ} \mathrm{C}\) as a result?

What mass of ice can be melted with the same quantity of heat as required to raise the temperature of \(3.50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}(1)\) by \(50.0^{\circ} \mathrm{C} ?\left[\Delta H_{\text {fusion }}^{\circ}=6.01 \mathrm{kJ} / \mathrm{mol}\right.\) \(\left.\mathrm{H}_{2} \mathrm{O}(\mathrm{s})\right]\)

Construct a concept map to show the use of enthalpy for chemical reactions.

Calculate the quantity of work associated with a \(3.5 \mathrm{L}\) expansion of a gas \((\Delta V)\) against a pressure of \(748 \space\mathrm{mmHg}\) in the units (a) atm \(\mathrm{L} ;\) (b) joules (J); (c) calories (cal).

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