/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 26 For the following processes, cal... [FREE SOLUTION] | 91影视

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Chapter 5: Problem 26

For the following processes, calculate the change in internal energy of the system and determine whether the process is endothermic or exothermic: (a) A balloon is heated by adding 850 J of heat. It expands, doing $382 \mathrm{~J}\( of work on the atmosphere. (b) A \)50-g$ sample of water is cooled from \(30^{\circ} \mathrm{C}\) to \(15^{\circ} \mathrm{C}\), thereby losing approximately \(3140 \mathrm{~J}\) of heat. (c) A chemical reaction releases \(6.47 \mathrm{~kJ}\) of heat and does no work on the surroundings.

Short Answer

Expert verified
(a) The change in internal energy is \(鈭哢 = 468 \mathrm{~J}\), and the process is endothermic. (b) The change in internal energy is \(鈭哢 = -3140 \mathrm{~J}\), and the process is exothermic. (c) The change in internal energy is \(鈭哢 = -6470 \mathrm{~J}\), and the process is exothermic.

Step by step solution

01

(a) Calculate the change in internal energy

: Given that 850 J of heat is added to the balloon (Q = 850 J) and the balloon expands, doing 382 J of work (W = 382 J). We can now find the change in internal energy using the previously discussed equation: 鈭哢 = Q - W 鈭哢 = 850 J - 382 J 鈭哢 = 468 J Since Q > 0, the process is endothermic.
02

(b) Calculate the change in internal energy of cooling water

: In this case, the 50-g sample of water loses 3140 J of heat (Q = -3140 J). There is no work done, so W = 0. We can find the change in internal energy using the same equation: 鈭哢 = Q - W 鈭哢 = -3140 J - 0 鈭哢 = -3140 J Since Q < 0, the process is exothermic.
03

(c) Calculate the change in internal energy of the chemical reaction

: For this problem, the chemical reaction releases 6.47 kJ of heat (Q = -6470 J, we use negatives since it's released). Since there is no work done on the surroundings (W = 0), we can find the change in internal energy using the same equation: 鈭哢 = Q - W 鈭哢 = -6470 J - 0 鈭哢 = -6470 J As Q < 0, the process is exothermic.

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Most popular questions from this chapter

The heat of combustion of fructose, \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\), is \(-2812 \mathrm{~kJ} / \mathrm{mol}\). If a fresh golden delicious apple weighing \(4.23 \mathrm{oz}(120 \mathrm{~g})\) contains \(16.0 \mathrm{~g}\) of fructose, what caloric content does the fructose contribute to the apple?

(a) A baseball weighs \(5.13 \mathrm{oz}\). What is the kinetic energy, in joules, of this baseball when it is thrown by a major-league pitcher at \(95.0 \mathrm{mi} / \mathrm{h}\) ? (b) By what factor will the kinetic energy change if the speed of the baseball is decreased to \(55.0 \mathrm{mi} / \mathrm{h}\) ? (c) What happens to the kinetic energy when the baseball is caught by the catcher?

(a) Calculate the kinetic energy, in joules, of a \(1200-\mathrm{kg}\) automobile moving at \(18 \mathrm{~m} / \mathrm{s}\). (b) Convert this energy to calories. (c) What happens to this energy when the automobile brakes to a stop?

When a \(6.50\)-g sample of solid sodium hydroxide dissolves in \(100.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure 5.17), the temperature rises from \(21.6\) to \(37.8^{\circ} \mathrm{C}\). (a) Calculate the quantity of heat (in \(\mathrm{kJ}\) ) released in the reaction. (b) Using your result from part (a), calculate \(\Delta H\) (in \(\mathrm{kJ} / \mathrm{mol} \mathrm{NaOH}\) ) for the solution process. Assume that the specific heat of the solution is the same as that of pure water.

(a) When a 4.25-g sample of solid ammonium nitrate dissolves in \(60.0 \mathrm{~g}\) of water in a coffee-cup calorimeter (Figure \(5.17)\), the temperature drops from \(22.0\) to \(16.9^{\circ} \mathrm{C}\). Calculate \(\Delta H\left(\right.\) in \(\mathrm{kJ} / \mathrm{mol} \mathrm{} \mathrm{NH}_{4} \mathrm{NO}_{3}\) ) for the solution process: $$ \mathrm{NH}_{4} \mathrm{NO}_{3}(s) \longrightarrow \mathrm{NH}_{4}^{+}(a q)+\mathrm{NO}_{3}^{-}(a q) $$ Assume that the specific heat of the solution is the same as that of pure water. (b) Is this process endothermic or exothermic?
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