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Chapter 6: Problem 130

Calculate the internal energy of a Goodyear blimp filled with helium gas at \(1.2 \times 10^{5} \mathrm{~Pa}\). The volume of the blimp is \(5.5 \times 10^{3} \mathrm{~m}^{3}\). If all the energy were used to heat 10.0 tons of copper at \(21^{\circ} \mathrm{C},\) calculate the final temperature of the metal. (Hint: See Section 5.7 for help in calculating the internal energy of a gas. 1 ton \(=9.072 \times 10^{5} \mathrm{~g} .\)

Short Answer

Expert verified
The final temperature of the copper will be the result calculated in step 3.

Step by step solution

01

Calculate Internal Energy of the Helium Gas

The internal energy for an ideal gas can be expressed as \( U = \frac{3}{2} PV \), where P is the pressure and V is the volume. The pressure is given as \(1.2 \times 10^5 Pa\) and the volume is \(5.5 \times 10^3 m^3\). Substituting these values into the equation, the internal energy is calculated as \( U = \frac{3}{2} \times 1.2 \times 10^5 Pa \times 5.5 \times 10^3 m^3 \).
02

Convert tons to grams

The problem provides a conversion for 1 ton to grams. Given the copper mass is 10 tons, it can be converted to grams by multiplying by the conversion factor \(10 tons \times 9.072 \times 10^5 g/ton\).
03

Calculate Final Temperature of Copper

Upon transfer of energy from the helium to the copper, the internal energy manifests as heat and the rise in temperature of copper can be given by \( \Delta T = \frac{Q}{mC} = \frac{U}{mC}\), where m is the mass of the copper, C is the specific heat capacity of copper (0.385 J/g°C), and U is the internal energy calculated in Step 1. Using the mass calculated in Step 2, the final temperature can be obtained by adding the initial temperature of copper to the change in temperature: \(T_{final} = T_{initial} + \Delta T\).

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Most popular questions from this chapter

The combustion of \(0.4196 \mathrm{~g}\) of a hydrocarbon releases \(17.55 \mathrm{~kJ}\) of heat. The masses of the products are \(\mathrm{CO}_{2}=1.419 \mathrm{~g}\) and \(\mathrm{H}_{2} \mathrm{O}=0.290 \mathrm{~g}\). (a) What is the empirical formula of the compound? (b) If the approximate molar mass of the compound is \(76 \mathrm{~g}\), calculate its standard enthalpy of formation.

The standard enthalpies of formation of ions in aqueous solutions are obtained by arbitrarily assigning a value of zero to \(\mathrm{H}^{+}\) ions; that is, \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}^{+}(a q)\right]=0\) (a) For the following reaction $$\begin{array}{r}\mathrm{HCl}(g) \stackrel{\mathrm{H}_{2} \mathrm{O}}{\longrightarrow} \mathrm{H}^{+}(aq)+\mathrm{Cl}^{-}(a q) \\\\\Delta H^{\circ}=-74.9 \mathrm{~kJ} / \mathrm{mol}\end{array}$$ calculate \(\Delta H_{\mathrm{f}}^{\circ}\) for the \(\mathrm{Cl}^{-}\) ions. (b) Given that \(\Delta H_{\mathrm{f}}^{\circ}\) for \(\mathrm{OH}^{-}\) ions is \(-229.6 \mathrm{~kJ} / \mathrm{mol}\), calculate the enthalpy of neutralization when 1 mole of a strong monoprotic acid (such as \(\mathrm{HCl}\) ) is titrated by 1 mole of a strong base (such as \(\mathrm{KOH}\) ) at \(25^{\circ} \mathrm{C}\).

State Hess's law. Explain, with one example, the usefulness of this law in thermochemistry.

Predict the value of \(\Delta H_{\mathrm{f}}^{\circ}\) (greater than, less than, or equal to zero) for these elements at \(25^{\circ} \mathrm{C}:\) (a) \(\mathrm{Br}_{2}(g)\) \(\mathrm{Br}_{2}(l)\) (b) \(\mathrm{I}_{2}(g) ; \mathrm{I}_{2}(s)\)

From the following data, $$ \begin{array}{c} \mathrm{C} \text { (graphite) }+\mathrm{O}_{2}(g) \longrightarrow \mathrm{CO}_{2}(g) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-393.5 \mathrm{~kJ} / \mathrm{mol} \\ \mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(g) \longrightarrow \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-285.8 \mathrm{~kJ} / \mathrm{mol} \\ 2 \mathrm{C}_{2} \mathrm{H}_{6}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \\ \Delta H_{\mathrm{rxn}}^{\circ}=-3119.6 \mathrm{~kJ} / \mathrm{mol} \end{array} $$ calculate the enthalpy change for the reaction $$ 2 \mathrm{C}(\text { graphite })+3 \mathrm{H}_{2}(g) \longrightarrow \mathrm{C}_{2} \mathrm{H}_{6}(g) $$
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