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Chapter 7: Problem 31

A 125 \(g\) stainless steel ball bearing \((\mathrm{spht}=\) \(0.50 \mathrm{Jg}^{-1}\) \(\left.^{\circ} \mathrm{C}^{-1}\right)\) at \(525^{\circ} \mathrm{C}\) is dropped into \(75.0 \mathrm{mL}\) of water at \(28.5^{\circ} \mathrm{C}\) in an open Styrofoam cup. As a result, the water is brought to a boil when the temperature reaches \(100.0^{\circ} \mathrm{C} .\) What mass of water vaporizes while the boiling continues? \(\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).

Short Answer

Expert verified
The mass of the water that vaporizes can be found by using the law of conservation of energy and the heat of vaporization of water.

Step by step solution

01

Calculate heat lost by the steel ball.

The first step is to calculate the heat (q) lost by the steel ball bearing as it cools from \(525^{\circ} C\) to \(100^{\circ} C\). The formula to use is \(q_{\text{steel}} = m_{\text{steel}} \cdot c_{\text{steel}} \cdot \Delta T_{\text{steel}}\), where \(m_{\text{steel}} = 125 \, g\) is the mass of the steel, \(c_{\text{steel}} = 0.50 \, J \, g^{-1} \, ^{\circ}C^{-1}\) is the specific heat of the steel, and \(\Delta T_{\text{steel}}\) is the change in temperature of the steel, which is \(525^{\circ} C - 100^{\circ} C = 425^{\circ} C\).
02

Calculate initial heat absorbed by water to reach boiling point.

Next, calculate the heat (q) absorbed by the water as it heats up from \(28.5^{\circ} C\) to \(100^{\circ} C\). The formula to use is \(q_{\text{water}} = m_{\text{water}} \cdot c_{\text{water}} \cdot \Delta T_{\text{water}}\), where \(m_{\text{water}} = 75.0 \, g\) (since 1mL of water has mass of 1g), \(c_{\text{water}} = 4.18 \, J \, g^{-1} \, ^{\circ}C^{-1}\), and \(\Delta T_{\text{water}}\) is the change in temperature of the water, which is \(100.0^{\circ} C - 28.5^{\circ} C = 71.5^{\circ} C\).
03

Find heat absorbed by water to vaporize.

The amount of heat (q) required to vaporize the water can be found from the difference of the heat lost by the steel and the heat initially absorbed by the water to reach the boiling point, i.e., \(q_{\text{vaporize}} = q_{\text{steel}} - q_{\text{water}}\).
04

Calculate mass of water vaporized.

Finally, find the mass of the water that vaporizes. This can be found by using the formula \(q_{\text{vaporize}} = m_{\text{vaporized}} \cdot \Delta H_{\text{vap}}\), where \(\Delta H_{\text{vap}} = 40.6 \, kJ/mol\) is the heat of vaporization of the water. The mass can be found by isolating it in the above formula: \(m_{\text{vaporized}} = \frac{q_{\text{vaporize}}}{\Delta H_{\text{vap}}}\). Note that \(\Delta H_{\text{vap}}\) needs to be converted to \(J \, g^{-1}\) to match the units with \(q_{\text{vaporize}}\).

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Most popular questions from this chapter

An alternative approach to bomb calorimetry is to establish the heat capacity of the calorimeter, exclusive of the water it contains. The heat absorbed by the water and by the rest of the calorimeter must be calculated separately and then added together. A bomb calorimeter assembly containing \(983.5 \mathrm{g}\) water is calibrated by the combustion of \(1.354 \mathrm{g}\) anthracene. The temperature of the calorimeter rises from 24.87 to \(35.63^{\circ} \mathrm{C} .\) When \(1.053 \mathrm{g}\) citric acid is burned in the same assembly, but with 968.6 g water, the temperature increases from 25.01 to \(27.19^{\circ} \mathrm{C}\). The heat of combustion of anthracene, \(\mathrm{C}_{14} \mathrm{H}_{10}(\mathrm{s}),\) is \(-7067 \mathrm{kJ} / \mathrm{mol}\) \(\mathrm{C}_{14} \mathrm{H}_{10} \cdot\) What is the heat of combustion of citric acid, \(\mathrm{C}_{6} \mathrm{H}_{8} \mathrm{O}_{7},\) expressed in \(\mathrm{kJ} / \mathrm{mol} ?\)

A 74.8 g sample of copper at \(143.2^{\circ} \mathrm{C}\) is added to an insulated vessel containing \(165 \mathrm{mL}\) of glycerol, \(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}(\mathrm{l})(d=1.26 \mathrm{g} / \mathrm{mL}),\) at \(24.8^{\circ} \mathrm{C} .\) The final temperature is \(31.1^{\circ} \mathrm{C}\). The specific heat of copper is \(0.385 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) What is the heat capacity of glycerol in \(\mathrm{Jmol}^{-1}\) \(^{\circ} \mathrm{C}^{-1} ?\)

Under the entry \(\mathrm{H}_{2} \mathrm{SO}_{4},\) a reference source lists many values for the standard enthalpy of formation. For example, for pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1), \Delta H_{\mathrm{f}}^{\circ}=-814.0 \mathrm{kJ} / \mathrm{mol}\) for a solution with \(1 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) per mole of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) \(-841.8 ;\) with \(10 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-880.5 ;\) with \(50 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-886.8 ;\) with \(100 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-887.7 ;\) with \(500 \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\) \(-890.5 ;\) with \(1000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-892.3 ;\) with \(10,000 \mathrm{mol}\) \(\mathrm{H}_{2} \mathrm{O},-900.8 ;\) and with \(100,000 \mathrm{mol} \mathrm{H}_{2} \mathrm{O},-907.3\) (a) Explain why these values are not all the same. (b) The value of \(\Delta H_{\mathrm{f}}^{\circ}\left[\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\right]\) in an infinitely dilute solution is \(-909.3 \mathrm{kJ} / \mathrm{mol} .\) What data from this chapter can you cite to confirm this value? Explain. (c) If \(500.0 \mathrm{mL}\) of \(1.00 \mathrm{M} \mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is prepared from pure \(\mathrm{H}_{2} \mathrm{SO}_{4}(1),\) what is the approximate change in temperature that should be observed? Assume that the \(\mathrm{H}_{2} \mathrm{SO}_{4}(1)\) and \(\mathrm{H}_{2} \mathrm{O}(1)\) are at the same temperature initially and that the specific heat of the \(\mathrm{H}_{2} \mathrm{SO}_{4}(\mathrm{aq})\) is about \(4.2 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1}\).

A 1.22 kg piece of iron at \(126.5^{\circ} \mathrm{C}\) is dropped into \(981 \mathrm{g}\) water at \(22.1^{\circ} \mathrm{C} .\) The temperature rises to \(34.4^{\circ} \mathrm{C} .\) What will be the final temperature if this same piece of iron at \(99.8^{\circ} \mathrm{C}\) is dropped into \(325 \mathrm{mL}\) of glycerol, \(\mathrm{HOCH}_{2} \mathrm{CH}(\mathrm{OH}) \mathrm{CH}_{2} \mathrm{OH}(1)\) at \(26.2^{\circ} \mathrm{C} ?\) For glycerol, \(d=1.26 \mathrm{g} / \mathrm{mL} ; C_{n}=219 \mathrm{JK}^{-1} \mathrm{mol}^{-1}\).

Construct a concept map to show the interrelationships between path-dependent and pathindependent quantities in thermodynamics.
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