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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Most popular questions from this chapter

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

The internal energy of a fixed quantity of an ideal gas depends only on its temperature. A sample of an ideal gas is allowed to expand at a constant temperature (isothermal expansion). (a) Does the gas do work? (b) Does the gas exchange heat with its surroundings? (c) What happens to the temperature of the gas? (d) What is \(\Delta U\) for the gas?

Use Hess's law to determine \(\Delta H^{\circ}\) for the reaction $$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}), \text { given that }$$ $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ &\left.\qquad \Delta H^{\circ}=-110.54 \mathrm{k} \mathrm{J}\right] \end{array}$$ $$\begin{aligned} &\text { C(graphite) }+\mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})\\\ &&\Delta H^{\circ}=-393.51 \mathrm{kJ} \end{aligned}$$

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Briefly describe each of the following ideas or methods: (a) law of conservation of energy; (b) bomb calorimetry; (c) function of state; (d) enthalpy diagram; (e) Hess's law.

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