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Chapter 6: Problem 92

The complete combustion of acetylene, \(\mathrm{C}_{2} \mathrm{H}_{2}(g)\), produces 1300\. kJ of energy per mole of acetylene consumed. How many grams of acetylene must be burned to produce enough heat to raise the temperature of \(1.00\) gal water by \(10.0^{\circ} \mathrm{C}\) if the process is \(80.0 \%\) efficient? Assume the density of water is \(1.00 \mathrm{~g} / \mathrm{cm}^{3}\)

Short Answer

Expert verified
Approximately 3.96 grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal of water by 10.0°C, considering an 80.0% efficiency of the process.

Step by step solution

01

Convert gallons of water to mass.

First, we need to convert 1.00 gal of water to grams. To do this, we will use the density of water, which is given as 1.00 g/cm³. We know that 1 gal = 3,785.41 cm³, so we can calculate the mass of water as follows: Mass of water = Volume × Density = 1.00 gal × (3,785.41 cm³/gal) × (1.00 g/cm³) = 3,785.41 g
02

Calculate the heat required to raise the water's temperature.

Next, we have to calculate the amount of heat needed to raise the temperature of the water by 10.0°C. We will use the specific heat capacity of water, which is 4.18 J/g°C. The heat can be calculated using the formula: Heat required (Q) = Mass × Specific heat × ΔTemperature Q = (3,785.41 g) × (4.18 J/g°C) × (10.0 °C) = 158,394.52 J Since we need to work with kJ, we must convert: Q = 158,394.52 J × (1 kJ / 1000 J) = 158.39 kJ
03

Account for efficiency.

As the process is 80.0% efficient, not all of the heat generated by burning acetylene will be used to heat the water. To find the actual heat needed, we can use the following formula: Actual heat required = Heat required / Process efficiency = 158.39 kJ / 0.8 = 198.00 kJ
04

Calculate the moles of acetylene needed.

Since each mole of acetylene generates 1300 kJ of energy when completely combusted, we can now calculate the number of moles needed to generate 198.00 kJ of heat: Moles of acetylene = Actual heat required / Heat per mole = 198.00 kJ / 1300 kJ/mol = 0.1523 mol
05

Calculate the mass of acetylene needed.

Finally, we can find the grams of acetylene required by using the molar mass of acetylene, which is 26.04 g/mol: Mass of acetylene = Moles × Molar mass = 0.1523 mol × 26.04 g/mol = 3.96 g So, approximately 3.96 grams of acetylene must be burned to produce enough heat to raise the temperature of 1.00 gal of water by 10.0°C, considering an 80.0% efficiency of the process.

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Most popular questions from this chapter

One mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) at \(1.00 \mathrm{~atm}\) and \(100 .^{\circ} \mathrm{C}\) occupies a volume of \(30.6 \mathrm{~L}\). When one mole of \(\mathrm{H}_{2} \mathrm{O}(g)\) is condensed to one mole of \(\mathrm{H}_{2} \mathrm{O}(l)\) at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}, 40.66 \mathrm{~kJ}\) of heat is released. If the density of \(\mathrm{H}_{2} \mathrm{O}(l)\) at this temperature and pressure is \(0.996 \mathrm{~g} / \mathrm{cm}^{3}\), calculate \(\Delta E\) for the condensation of one mole of water at \(1.00 \mathrm{~atm}\) and \(100 .{ }^{\circ} \mathrm{C}\).

Acetylene \(\left(\mathrm{C}_{2} \mathrm{H}_{2}\right)\) and butane \(\left(\mathrm{C}_{4} \mathrm{H}_{10}\right)\) are gaseous fuels with enthalpies of combustion of \(-49.9 \mathrm{~kJ} / \mathrm{g}\) and \(-49.5 \mathrm{~kJ} / \mathrm{g}\), respectively. Compare the energy available from the combustion of a given volume of acetylene to the combustion energy from the same volume of butane at the same temperature and pressure.

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.

Consider the dissolution of \(\mathrm{CaCl}_{2}\) : $$ \mathrm{CaCl}_{2}(s) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=-81.5 \mathrm{~kJ} $$ An \(11.0-\mathrm{g}\) sample of \(\mathrm{CaCl}_{2}\) is dissolved in \(125 \mathrm{~g}\) water, with both substances at \(25.0^{\circ} \mathrm{C}\). Calculate the final temperature of the solution assuming no heat loss to the surroundings and assuming the solution has a specific heat capacity of \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\).

Write reactions for which the enthalpy change will be a. \(\Delta H_{\mathrm{f}}^{\circ}\) for solid aluminum oxide. b. The standard enthalpy of combustion of liquid ethanol, \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(l) .\) c. The standard enthalpy of neutralization of sodium hydroxide solution by hydrochloric acid. d. \(\Delta H_{\mathrm{f}}^{\circ}\) for gaseous vinyl chloride, \(\mathrm{C}_{2} \mathrm{H}_{3} \mathrm{Cl}(g)\). e. The enthalpy of combustion of liquid benzene, \(\mathrm{C}_{6} \mathrm{H}_{6}(l)\). f. The enthalpy of solution of solid ammonium bromide.
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