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Chapter 6: Problem 65

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 \(4.18 \mathrm{~J} /{ }^{\circ} \mathrm{C} \cdot \mathrm{g}\)

Short Answer

Expert verified
The final temperature of the solution after dissolving an 11.0-g sample of CaCl2 in 125 g of water, both initially at 25.0°C, is approximately 40.4°C.

Step by step solution

01

To find the moles of CaCl2, use the mass and molar mass as follows: moles of CaCl2 = (mass of CaCl2) / (molar mass of CaCl2) The molar mass of CaCl2 = (40.08 + 2 × 35.45)g/mol = 110.98 g/mol moles of CaCl2 = (11.0 g) / (110.98 g/mol) = 0.0991 mol #step 2: Calculate the heat produced during dissolution#

Using the enthalpy change of dissolution, ΔH, calculate the heat produced when CaCl2 dissolves: Heat produced (q) = moles of CaCl2 × ΔH q = 0.0991 mol × (-81.5 kJ/mol) = -8.075 kJ Since the heat produced is negative, it means the dissolution process is exothermic and heat is released. #step 3: Convert heat produced to Joules#
02

Convert the heat produced from kJ to J: q = -8.075 kJ × (1000 J/1 kJ) = -8075 J #step 4: Calculate the heat absorbed by the water#

When heat is neither lost nor gained, the heat absorbed by the water equals the heat produced during dissolution: Heat absorbed by water (qw) = -q = 8075 J #step 5: Calculate the temperature change of the water#
03

Use the specific heat formula to find the temperature change of the water: qw = mass of water × specific heat capacity × ΔT where ΔT is the temperature change. Rearrange the formula to solve for ΔT: ΔT = qw / (mass of water × specific heat capacity) ΔT = (8075 J) / (125 g × 4.18 J/ (°C·g) ) ≈ 15.4 °C #step 6: Calculate the final temperature of the solution#

To find the final temperature of the solution, add the temperature change to the initial temperature: T_final = T_initial + ΔT T_final = 25.0 °C + 15.4 °C ≈ 40.4 °C The final temperature of the solution is approximately 40.4 °C.

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Most popular questions from this chapter

Are the following processes exothermic or endothermic? a. When solid \(\mathrm{KBr}\) is dissolved in water, the solution gets colder. b. Natural gas \(\left(\mathrm{CH}_{4}\right)\) is burned in a furnace. c. When concentrated \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is added to water, the solution gets very hot. d. Water is boiled in a teakettle.

In which of the following systems is(are) work done by the surroundings on the system? Assume pressure and temperature are constant. a. \(2 \mathrm{SO}_{2}(g)+\mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{SO}_{3}(g)\) b. \(\mathrm{CO}_{2}(s) \longrightarrow \mathrm{CO}_{2}(g)\) c. \(4 \mathrm{NH}_{3}(g)+7 \mathrm{O}_{2}(g) \longrightarrow 4 \mathrm{NO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)\) d. \(\mathrm{N}_{2} \mathrm{O}_{4}(g) \longrightarrow 2 \mathrm{NO}_{2}(g)\) e. \(\mathrm{CaCO}_{3}(s) \longrightarrow \mathrm{CaCO}(s)+\mathrm{CO}_{2}(g)\)

Calculate \(\Delta E\) for each of the following. a. \(q=-47 \mathrm{~kJ}, w=+88 \mathrm{~kJ}\) b. \(q=+82 \mathrm{~kJ}, w=-47 \mathrm{~kJ}\) c. \(q=+47 \mathrm{~kJ}, w=0\) d. In which of these cases do the surroundings do work on the system?

In a bomb calorimeter, the reaction vessel is surrounded by water that must be added for each experiment. Since the amount of water is not constant from experiment to experiment, the mass of water must be measured in each case. The heat capacity of the calorimeter is broken down into two parts: the water and the calorimeter components. If a calorimeter contains \(1.00 \mathrm{~kg}\) water and has a total heat capacity of \(10.84 \mathrm{~kJ} /{ }^{\circ} \mathrm{C}\), what is the heat capacity of the calorimeter components?

Hydrogen gives off \(120 . \mathrm{J} / \mathrm{g}\) of energy when burned in oxygen, and methane gives off \(50 . \mathrm{J} / \mathrm{g}\) under the same circumstances. If a mixture of \(5.0 \mathrm{~g}\) hydrogen and \(10 . \mathrm{g}\) methane is burned, and the heat released is transferred to \(50.0 \mathrm{~g}\) water at \(25.0^{\circ} \mathrm{C}\), what final temperature will be reached by the water?
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