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Chapter 6: Problem 29

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{~kJ}, w=+88 \mathrm{~kJ}\) b. \(q=+82 \mathrm{~kJ}, w=-47 \mathrm{~kJ}\) c. \(q=+47 \mathrm{~kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

Short Answer

Expert verified
The change in internal energy (\(\Delta E\)) for each case is calculated as follows: a. \(\Delta E = 41 \mathrm{~kJ}\) b. \(\Delta E = 35 \mathrm{~kJ}\) c. \(\Delta E = 47 \mathrm{~kJ}\) The surroundings do work on the system only in case a.

Step by step solution

01

Case a: Calculate \(\Delta E\)

To calculate the change in internal energy for case a, we will plug in the given values of \(q\) and \(w\) into the formula: \[\Delta E = (-47 \mathrm{~kJ}) + (+88 \mathrm{~kJ})\] Now, we simply add the two values together: \[\Delta E = 41 \mathrm{~kJ}\]
02

Case b: Calculate \(\Delta E\)

For case b, plug in the given values of \(q\) and \(w\) into the formula: \[\Delta E = (+82 \mathrm{~kJ}) + (-47 \mathrm{~kJ})\] Now, add the two values together: \[\Delta E = 35 \mathrm{~kJ}\]
03

Case c: Calculate \(\Delta E\)

For case c, plug in the given values of \(q\) and \(w\) into the formula: \[\Delta E = (+47 \mathrm{~kJ}) + (0)\] Since work done (\(w\)) is zero, the change in internal energy is equal to the heat added to the system: \[\Delta E = 47 \mathrm{~kJ}\]
04

Identify cases where surroundings do work on the system

The surroundings do work on the system when the work (\(w\)) is positive. Looking at the given values for each case, we find that: - In case a, \(w = +88\) kJ (positive) - In case b, \(w = -47\) kJ (negative) - In case c, \(w = 0\) kJ (zero) Thus, the surroundings do work on the system only in case a.

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Most popular questions from this chapter

The enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(l)\) is formed is \(-891 \mathrm{~kJ} / \mathrm{mol}\) and the enthalpy of combustion of \(\mathrm{CH}_{4}(\mathrm{~g})\) when \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g})\) is formed is \(-803 \mathrm{~kJ} / \mathrm{mol} .\) Use these data and Hess's law to determine the enthalpy of vaporization for water.

The specific heat capacity of silver is \(0.24 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\). a. Calculate the energy required to raise the temperature of \(150.0 \mathrm{~g}\) Ag from \(273 \mathrm{~K}\) to \(298 \mathrm{~K}\). b. Calculate the energy required to raise the temperature of \(1.0 \mathrm{~mol} \mathrm{Ag}\) by \(1.0^{\circ} \mathrm{C}\) (called the molar heat capacity of silver). c. It takes \(1.25 \mathrm{~kJ}\) of energy to heat a sample of pure silver from \(12.0^{\circ} \mathrm{C}\) to \(15.2^{\circ} \mathrm{C}\). Calculate the mass of the sample of silver.

The sun supplies energy at a rate of about \(1.0\) kilowatt per square meter of surface area ( 1 watt \(=1 \mathrm{~J} / \mathrm{s}\) ). The plants in an agricultural field produce the equivalent of \(20 . \mathrm{kg}\) sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) per hour per hectare ( \(1 \mathrm{ha}=10,000 \mathrm{~m}^{2}\) ). Assuming that sucrose is produced by the reaction \(12 \mathrm{CO}_{2}(g)+11 \mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}(s)+12 \mathrm{O}_{2}(g)\) \(\Delta H=5640 \mathrm{~kJ}\) calculate the percentage of sunlight used to produce the sucrosethat is, determine the efficiency of photosynthesis.

It takes \(585 \mathrm{~J}\) of energy to raise the temperature of \(125.6 \mathrm{~g}\) mercury from \(20.0^{\circ} \mathrm{C}\) to \(53.5^{\circ} \mathrm{C}\). Calculate the specific heat capacity and the molar heat capacity of mercury.

The enthalpy change for a reaction is a state function and it is an extensive property. Explain.
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