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Chapter 7: Problem 9

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 \(\mathrm{Jmol}^{-1}\) \(^{\circ} \mathrm{C}^{-1} ?\)

Short Answer

Expert verified
The specific heat capacity of glycerol is 192.5 J/mol°C.

Step by step solution

01

Calculate the Heat Lost by the Copper

We use the formula \(q = mc\Delta T\) to calculate the amount of heat lost by the copper. Here, m is the mass of copper which is 74.8 grams, ΔT (change in temperature) is \(143.2 - 31.1 = 112.1°C\) and c is the specific heat capacity of copper which is 0.385 \(J/g°C\). The heat lost by the copper is thus \(-74.8 * 0.385 * 112.1 = -3293.8J\). The sign is negative because heat is lost by the system.
02

Calculate the Heat Gained by the Glycerol

Given that in an isolated system, the total amount of heat stays constant, the heat gained by glycerol equals to lost heat by copper. So, the quantity of heat gained by the glycerol is \(3293.8 J\).
03

Calculate the Mass of Glycerol

To find the mass of glycerol, we multiply the volume by the density. The volume is given as 165 mL and the density as 1.26 \(g/mL\). Thus, the mass of glycerol is \(165 * 1.26 = 207.9g\).
04

Find the specific heat capacity of Glycerol

Using the heat formula \(q = mc\Delta T\) backwards, we find the specific heat capacity c by isolating c in the equation \(c = q / (m * \Delta T)\). Hence, the specific heat capacity c of glycerol will be \(3293.8 / (207.9 * (31.1 - 24.8)) = 2.09 J/g°C\).
05

Convert the specific heat to J/mol°C

Convert the specific heat of Glycerol from \(J/g°C\) to \(J/mol°C\) using the molecular weight of Glycerol. The molar mass of glycerol is 92.09 g/mol, thus, the molar specific heat capacity is \(2.09 * 92.09 = 192.5 J/mol°C\).

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Most popular questions from this chapter

Substitute natural gas (SNG) is a gaseous mixture containing \(\mathrm{CH}_{4}(\mathrm{g})\) that can be used as a fuel. One reaction for the production of SNG is $$\begin{aligned} 4 \mathrm{CO}(\mathrm{g})+8 \mathrm{H}_{2}(\mathrm{g}) & \longrightarrow \\ 3 \mathrm{CH}_{4}(\mathrm{g})+\mathrm{CO}_{2}(\mathrm{g})+2 \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) & \Delta H^{\circ}=? \end{aligned}$$ Use appropriate data from the following list to determine \(\Delta H^{\circ}\) for this SNG reaction. $$\begin{array}{l} \text { C(graphite) }+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}(\mathrm{g}) \\ \quad\quad\quad\quad\quad\quad\quad\quad\qquad \Delta H^{\circ}=-110.5 \mathrm{k} \mathrm{J} \end{array}$$$$\mathrm{CO}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g}) \quad \Delta H^{\circ}=-283.0 \mathrm{kJ}$$ $$\mathrm{H}_{2}(\mathrm{g})+\frac{1}{2} \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \quad \Delta H^{\circ}=-285.8 \mathrm{kJ}$$ $$\begin{array}{l} \text { C(graphite) }+2 \mathrm{H}_{2}(\mathrm{g}) \longrightarrow \mathrm{CH}_{4}(\mathrm{g}) \\ \qquad \Delta H^{\circ}=-74.81 \mathrm{kJ} \end{array}$$ $$\begin{aligned} \mathrm{CH}_{4}(\mathrm{g})+2 \mathrm{O}_{2}(\mathrm{g}) \longrightarrow \mathrm{CO}_{2}(\mathrm{g})+& 2 \mathrm{H}_{2} \mathrm{O}(1) \\ & \Delta H^{\circ}=-890.3 \mathrm{kJ} \end{aligned}$$

Thermite mixtures are used for certain types of welding, and the thermite reaction is highly exothermic. $$\begin{array}{r} \mathrm{Fe}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+2 \mathrm{Fe}(\mathrm{s}) \\ \Delta H^{\circ}=-852 \mathrm{kJ} \end{array}$$ \(1.00 \mathrm{mol}\) of granular \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) and \(2.00 \mathrm{mol}\) of granular Al are mixed at room temperature \(\left(25^{\circ} \mathrm{C}\right),\) and a reaction is initiated. The liberated heat is retained within the products, whose combined specific heat over a broad temperature range is about \(0.8 \mathrm{Jg}^{-1}\) \(^{\circ} \mathrm{C}^{-1} .\) (The melting point of iron is \(1530^{\circ} \mathrm{C} .\) ) Show that the quantity of heat liberated is more than sufficient to raise the temperature of the products to the melting point of iron.

A 1.103 g sample of a gaseous carbon-hydrogenoxygen compound that occupies a volume of \(582 \mathrm{mL}\) at 765.5 Torr and \(25.00^{\circ} \mathrm{C}\) is burned in an excess of \(\mathrm{O}_{2}(\mathrm{g})\) in a bomb calorimeter. The products of the combustion are \(2.108 \mathrm{g} \mathrm{CO}_{2}(\mathrm{g}), 1.294 \mathrm{g} \mathrm{H}_{2} \mathrm{O}(1),\) and enough heat to raise the temperature of the calorimeter assembly from 25.00 to \(31.94^{\circ} \mathrm{C}\). The heat capacity of the calorimeter is \(5.015 \mathrm{kJ} /^{\circ} \mathrm{C}\). Write an equation for the combustion reaction, and indicate \(\Delta H^{\circ}\) for this reaction at \(25.00^{\circ} \mathrm{C}\).

A plausible final temperature when \(75.0 \mathrm{mL}\) of water at \(80.0^{\circ} \mathrm{C}\) is added to \(100.0 \mathrm{mL}\) of water at \(20^{\circ} \mathrm{C}\) is (a) \(28^{\circ} \mathrm{C} ;\) (b) \(40^{\circ} \mathrm{C} ;\) (c) \(46^{\circ} \mathrm{C} ;\) (d) \(50^{\circ} \mathrm{C}\)

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 \(25.0^{\circ} \mathrm{C} ?\left(\Delta H_{\mathrm{vap}}^{\circ}=40.6 \mathrm{kJ} / \mathrm{mol} \mathrm{H}_{2} \mathrm{O}\right)\).
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