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Chapter 7: Problem 30

What will be the final temperature of the water in an insulated container as the result of passing \(5.00 \mathrm{g}\) of steam, \(\mathrm{H}_{2} \mathrm{O}(\mathrm{g}),\) at \(100.0^{\circ} \mathrm{C}\) into \(100.0 \mathrm{g}\) of water at \(25.0^{\circ} \mathrm{C} ?\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

Short Answer

Expert verified
The final temperature of the water will be 52 °C.

Step by step solution

01

Calculate moles of steam

First, convert the mass of steam to moles using the molar mass of water (18.02 g/mol). For example, \(5.00 g \div 18.02 g/mol = 0.277 mol\)
02

Calculate heat released by steam

Next, calculate the heat released when the steam condenses using the molar enthalpy of vaporization: \(q = n \cdot \Delta H_{vap} = 0.277 mol \times 40.6 kJ/mol = 11.3 kJ\) This is the amount of heat that the steam releases as it cools and condenses.
03

Calculate the heat absorbed by the water and the final temperature

The heat absorbed by the 100.0 g of water will be the same as the heat released by the steam (due to conservation of energy in an insulated container), so we can use the specific heat equation \(q = m \cdot c \cdot \Delta T\) to solve for the change in temperature: \(11.3 kJ = 100.0 g \times 4.18 \times 10^{-3} kJ/g°C \times \Delta T \) . Solving for delta T, we get \( \Delta T = 27 °C \). Thus, the final temperature of the water will be the initial temperature plus the change in temperature: \(T_final = T_initial + \Delta T = 25 °C + 27 °C = 52 °C\).

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Most popular questions from this chapter

The standard heats of combustion \(\left(\Delta H^{\circ}\right)\) per mole of 1,3-butadiene, \(\mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g}) ;\) butane, \(\mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}) ;\) and \(\mathrm{H}_{2}(\mathrm{g})\) are \(-2540.2,-2877.6,\) and \(-285.8 \mathrm{kJ},\) respectively. Use these data to calculate the heat of hydrogenation of 1,3-butadiene to butane. $$\mathrm{C}_{4} \mathrm{H}_{6}(\mathrm{g})+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{C}_{4} \mathrm{H}_{10}(\mathrm{g}) \quad \Delta H^{\circ}=?$$ [Hint: Write equations for the combustion reactions. In each combustion, the products are \(\mathrm{CO}_{2}(\mathrm{g})\) and \(\left.\mathrm{H}_{2} \mathrm{O}(1) .\right]\)

A 1.397 g sample of thymol, \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O}(\mathrm{s})\) (a preservative and a mold and mildew preventative), is burned in a bomb calorimeter assembly. The temperature increase is \(11.23^{\circ} \mathrm{C},\) and the heat capacity of the bomb calorimeter is \(4.68 \mathrm{kJ} /^{\circ} \mathrm{C}\). What is the heat of combustion of thymol, expressed in kilojoules per mole of \(\mathrm{C}_{10} \mathrm{H}_{14} \mathrm{O} ?\)

A \(7.26 \mathrm{kg}\) shot (as used in the sporting event, the shot put) is dropped from the top of a building \(168 \mathrm{m}\) high. What is the maximum temperature increase that could occur in the shot? Assume a specific heat of \(0.47 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\) for the shot. Why would the actual measured temperature increase likely be less than the calculated value?

A \(1.00 \mathrm{g}\) sample of \(\mathrm{Ne}(\mathrm{g})\) at 1 atm pressure and \(27^{\circ} \mathrm{C}\) is allowed to expand into an evacuated vessel of \(2.50 \mathrm{L}\) volume. Does the gas do work? Explain.

A 1.620 g sample of naphthalene, \(C_{10} \mathrm{H}_{8}(\mathrm{s}),\) is completely burned in a bomb calorimeter assembly and a temperature increase of \(8.44^{\circ} \mathrm{C}\) is noted. If the heat of combustion of naphthalene is \(-5156 \mathrm{kJ} / \mathrm{mol} \mathrm{C}_{10} \mathrm{H}_{8}\) what is the heat capacity of the bomb calorimeter?
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