OpenStax Chemistry 2e Chapter 5 Section 2 - HedgeDoc

<style> /* Width of text in slides */ .slides { width: 1200px !important; } /* line space bullet list */ .reveal .slides section, .reveal .slides section > section { line-height: 1.5; } /* headers */ .reveal h1 { margin-bottom: 1em; font-weight: bold; font-size: 32pt } .reveal h2 { margin-bottom: 1em; font-weight: bold; font-size: 30pt } .reveal h3 { margin-bottom: 1em; font-weight: bold; font-size: 28pt } .reveal h4 { margin-bottom: 1em; font-weight: bold; font-size: 26pt } .reveal h5 { margin-bottom: 1em; font-weight: bold; font-size: 24pt } .reveal h6 { margin-bottom: 1em; font-weight: normal; font-size: 24pt } /* text */ .reveal { font-size: 24pt; font-weight: normal; } /* logo as background */ .container { background: no-repeat url(https://www.mcc.edu/MottStrong/images/mott-strong-3c-logo.png) top 20px left 10px; background-size: 5% !important; } code {color: #000000} .reveal table th { font-size: 0.80em} .reveal table td { font-size: 0.70em} /* Remove Border arround images */ .reveal section img { background:none; border:none; box-shadow:none; object-fit: contain; margin: 0; padding: 0; max-width: 100%; } .reveal figure img { box-shadow:none; background:none; border:none; max-width: 100%; object-fit: contain; } </style> $\require{mhchem}$ $\require{cancel}$ OpenStax Chemistry 2e Chapter 5 Section 2 === Michael Stogsdill Mott Community College --- ## Learning Objectives - Explain the technique of calorimetry - Calculate and interpret heat and related properties using typical calorimetry data --- ## Calorimetry - **Calorimetry** is a technique used to measure the amount of *heat* absorbed or released by a *chemical* or *physical process*. - The measurement of heat transfer using this approach requires defining a **system** and its **surroundings**. - The **system** is the substance or substances undergoing the chemical or physical change - The **surroundings** is all other matter, including components of the measurement apparatus, that either provide heat to the system or absorb heat from the system. --- ![Two diagrams labeled a and b are shown. Each is made up of two rectangular containers with a thermometer inserted into the top right and extending inside. There is a right facing arrow connecting each box in each diagram. The left container in diagram a depicts a pink and green swirling solution with the terms “Exothermic process” and “System” written in the center with arrows facing away from the terms pointing to “q.” The labels “Solution” and “Surroundings” are written at the bottom of the container. The right container in diagram a has the term “Solution” written at the bottom of the container and a red arrow facing up near the thermometer with the phrase “Temperature increased” next to it. The pink and green swirls are more blended in this container. The left container in diagram b depicts a purple and blue swirling solution with the terms “Endothermic process” and “System” written in the center with arrows facing away from the terms and “Solution” and “Surroundings” written at the bottom. The arrows point away from the letter “q.” The right container in diagram b has the term “Solution” written at the bottom and a red arrow facing down near the thermometer with the phrase “Temperature decreased” next to it. The blue and purple swirls are more blended in this container.](https://openstax.org/apps/archive/20250210.163914/resources/68b9d7102438342c6cf948437ecbe44f09267984) Note: **Figure 5.11** In a calorimetric determination, either (a) an exothermic process occurs and heat, q, is negative, indicating that thermal energy is transferred from the system to its surroundings, or (b) an endothermic process occurs and heat, q, is positive, indicating that thermal energy is transferred from the surroundings to the system. --- ### Simple Calorimeter <img src="https://openstax.org/apps/archive/20250210.163914/resources/d70f272ca875c80a369035a3d654a2b94778a9cf" alt="Two Styrofoam cups are shown nested in one another with a cover over the top. A thermometer and stirring rod are inserted through the cover and into the solution inside the cup, which is shown as a cut-away. The stirring rod has a double headed arrow next to it facing up and down. The liquid mixture inside the cup is labeled “Reaction mixture.”" style="zoom: 50%;" /> Note: **Figure 5.12** A simple calorimeter can be constructed from two polystyrene cups. A thermometer and stirrer extend through the cover into the reaction mixture. --- ### Laboratory Calorimeters <img src="https://openstax.org/apps/archive/20250210.163914/resources/b12a65603359caf431fd933a12fa1eefe4551359" alt="Two diagrams are shown and labeled a and b. Diagram a depicts a thermometer which passes through a disk-like insulating cover and into a metal cylinder which is labeled “metal inner vessel,” which is in turn nested in a metal cylinder labeled “metal outer vessel.” The inner cylinder rests on an insulating support ring. A stirrer passes through the insulating cover and into the inner cylinder as well. Diagram b shows an inner metal vessel half full of liquid resting on an insulating support ring and nested in a metal outer vessel. A precision temperature probe and motorized stirring rod are placed into the solution in the inner vessel and connected by wires to equipment exterior to the set-up." style="zoom:65%;" /> Note: **Figure 5.13** Commercial solution calorimeters range from (a) simple, inexpensive models for student use to (b) expensive, more accurate models for industry and research. --- ### Bomb Calorimeter <img src="https://doc.chemnotes.org/uploads/e9fe64af-e351-436b-a5c0-0e51d2d4a622.png" style="zoom:75%;" /> --- ### Calorimetry Calculations Calorimetry restricts heat transfer except to and from the reaction $\left( q_\textsf{rxn} \, \right)$ and the calorimeter's other contents $\left( q_\textsf{surr} \, \right)$. $$ q_\textsf{rxn} + q_\textsf{surr} = 0 $$ $$ q_\textsf{rxn} = - q_\textsf{surr} $$ $$ m_\textsf{rxn} \, c_\textsf{rxn} \, \Delta T_\textsf{rxn} \, = - m_\textsf{surr} \, c_\textsf{surr} \, \Delta T_\textsf{surr} $$ $$ m_\textsf{rxn} \, c_\textsf{rxn} \, \left( \Delta T^f_\textsf{rxn} \, - \, \Delta T^i_\textsf{rxn} \right) = - m_\textsf{surr} \, c_\textsf{surr} \, \left( \Delta T^f_\textsf{surr} \, - \, \Delta T^i_\textsf{surr} \right) $$ --- ## Question 18 <p style="text-align:left">How many milliliters of water at 23 °C with a density of 1.00 g/mL must be mixed with 180 mL (about 6 oz) of coffee at 95 °C so that the resulting combination will have a temperature of 60 °C? Assume that coffee and water have the same density and the same specific heat.</p> ---- Gather our variables. $c_{\textsf{water}} = 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  $\rho = 1.00 \, \frac{\textsf{g}}{\textsf{mL}}$  $V_\textsf{water} = ?$  $T^i_\textsf{water} = 23 ^\circ \textsf{C}$  $V_\textsf{coffee} = 180 \, \textsf{mL}$  $T^i_\textsf{coffee} = 95 ^\circ \textsf{C}$  $T^f = 60 ^\circ \textsf{C}$  ---- Calculate the temperature changes <br/> $\Delta T_\textsf{water} = T^f - T^i_\textsf{water} = 60 ^\circ \textsf{C} - 23 ^\circ \textsf{C} = 37 ^\circ \textsf{C}$  <br/> $\Delta T_\textsf{coffee} = T^f - T^i_\textsf{coffee} = 60 ^\circ \textsf{C} - 95 ^\circ \textsf{C} = -35 ^\circ \textsf{C}$  ---- The coffee cup is acting as a simple calorimeter. Once mixed, the heat gained by the water comes from the coffee. No heat leaves the cup. <br/> $\large q_\textsf{coffee} \hspace{2mm} + q_\textsf{water} \hspace{2mm} = 0$  $\large q_\textsf{coffee} \hspace{2mm} = - q_\textsf{water}$  <br/> $\large m_\textsf{coffee} \hspace{2mm} c \, \Delta T_\textsf{coffee} \hspace{2mm} = - m_\textsf{water} \hspace{2mm} c \, \Delta T_\textsf{water}$  ---- The mass of water and coffee can be replaced by the volume of these liquids times there volume. $\large V_\textsf{coffee} \hspace{2mm} \rho c \, \Delta T_\textsf{coffee} \hspace{2mm} = - V_\textsf{water} \hspace{2mm} \rho c \, \Delta T_\textsf{water}$  <br/> Because the water and coffee have the same *specific heat capacity* and density, these terms cancel.  $\large V_\textsf{coffee} \hspace{2mm} \Delta T_\textsf{coffee} \hspace{2mm} = - V_\textsf{water} \hspace{2mm} \Delta T_\textsf{water}$  <br/> The volume of water is therefore  $\large V_\textsf{water} = - V_\textsf{coffee} \hspace{2mm} \left( \displaystyle\frac{\Delta T_\textsf{coffee}}{\Delta T_\textsf{water}} \right)$  ---- Plugging in our values... <br/> $$ \large V_\textsf{water} = - 180 \, \textsf{mL} \hspace{2mm} \left( \displaystyle\frac{-35 ^\circ \textsf{C}}{37 ^\circ \textsf{C}} \right) = 170 \, \textsf{mL} $$  --- ## Question 19 <p style="text-align:left">How much will the temperature of a cup (180 g) of coffee at 95 °C be reduced when a 45 g silver spoon (specific heat 0.24 J/g °C) at 25 °C is placed in the coffee and the two are allowed to reach the same temperature? Assume that the coffee has the same density and specific heat as water.</p> ---- First, we gather our variables. $m_{\textsf{coffee}} = 180 \, \textsf{g}$  $m_{\textsf{spoon}} = 45 \, \textsf{g}$  $c_{\textsf{water}} = 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  $c_{\textsf{silver}} = 0.24 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  $T_{i}^{\textsf{coffee}} = 95 \, \textsf{C}$  $T_{i}^{\textsf{spoon}} = 25 \, \textsf{C}$  $T_{f} = ?$  ---- <p style="text-align:left">We start by assuming that no heat leaves the system. This means that all of the heat that leaves the coffee enters into the spoon. The total heat of the system is zero.</p> <br/> $q_{\textsf{spoon}} + q_{\textsf{coffee}} \, = 0$  $q_{\textsf{spoon}} = -q_{\textsf{coffee}}$  <br/> $\large m_{\textsf{spoon}}c_{\textsf{silver}} \left( T_{f} - T_{i}^{\textsf{spoon}} \right) = -m_{\textsf{coffee}}c_{\textsf{water}} \left(T_{f} - T_{i}^{\textsf{coffee}} \right)$  ---- Now we distribute across the subtractive terms. $$ \large m_{\textsf{spoon}}c_{\textsf{silver}}T_{f} - m_{\textsf{spoon}}c_{\textsf{silver}}T_{i}^{\textsf{spoon}} = m_{\textsf{coffee}}c_{\textsf{water}}T_{i}^{\textsf{coffee}} - m_{\textsf{coffee}}c_{\textsf{water}}T_{f} $$  <br/> Gather like terms.  $$ \large m_{\textsf{spoon}}c_{\textsf{silver}}T_{f} + m_{\textsf{coffee}}c_{\textsf{water}}T_{f} = m_{\textsf{coffee}}c_{\textsf{water}}T_{i}^{\textsf{coffee}} + m_{\textsf{spoon}}c_{\textsf{silver}}T_{i}^{\textsf{spoon}} $$  ---- We can now factor out the final temperature. $\large T_{f}(m_{\textsf{spoon}}c_{\textsf{silver}} + m_{\textsf{coffee}}c_{\textsf{water}}) = m_{\textsf{coffee}}c_{\textsf{water}}T_{i}^{\textsf{coffee}} + m_{\textsf{spoon}}c_{\textsf{silver}}T_{i}^{\textsf{spoon}}$  <br/> Finally, we can solve for the final temperature.  $$ \large T_{f} = \frac{m_{\textsf{coffee}}c_{\textsf{water}}T_{i}^{\textsf{coffee}} + m_{\textsf{spoon}}c_{\textsf{silver}}T_{i}^{\textsf{spoon}}}{m_{\textsf{spoon}}c_{\textsf{silver}} + m_{\textsf{coffee}}c_{\textsf{water}}} $$  ---- ...and plug in our values. <br/> $$ \large T_{f} = \frac{\left(180 \, \textsf{g}\right)\left(4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}\right)\left(95 \, \textsf{C}\right) + \left(45 \, \textsf{g}\right)\left(0.24 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}\right)\left(25 \, \textsf{C}\right)}{\left(45 \, \textsf{g}\right)\left(0.24 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}\right) + \left(180 \, \textsf{g}\right)\left(4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}\right)} $$  $$ \large= 94 \, ^\circ\textsf{C} $$  --- ## Question 22 <p style="text-align:left">A 70.0‑g piece of metal at 80.0 °C is placed in 100 g of water at 22.0 °C contained in a calorimeter like that shown in Figure 5.12. The metal and water come to the same temperature at 24.6 °C. How much heat did the metal give up to the water? What is the specific heat of the metal?</p> ---- Gather our variables $m_\textsf{metal} = 70.0 \, \textsf{g}$  $T^i_\textsf{metal} = 80.0 ^\circ \textsf{C}$  $c_{\textsf{metal}} = ?$  $m_\textsf{water} = 100 \, \textsf{g}$  $T^i_\textsf{water} = 22.0 ^\circ \textsf{C}$  $c_{\textsf{water}} = 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  $T^f = 24.6 ^\circ \textsf{C}$  ---- Calculate the temperature changes <br/> $\Delta T_\textsf{metal} = T^f - T^i_\textsf{metal} = 24.6 ^\circ \textsf{C} - 80.0 ^\circ \textsf{C} = -55.4 ^\circ \textsf{C}$  <br/> $\Delta T_\textsf{water} = T^f - T^i_\textsf{water} = 24.6 ^\circ \textsf{C} - 22.0 ^\circ \textsf{C} = 2.6 ^\circ \textsf{C}$  ---- The *heat* lost by the metal was gained by the water. <br/> $\large q_\textsf{metal} \hspace{2mm} = - q_\textsf{water}$  $\large m_\textsf{metal} \hspace{2mm} c_\textsf{metal} \hspace{2mm} \Delta T_\textsf{metal} \hspace{2mm} = - m_\textsf{water} \hspace{2mm} c_\textsf{water} \hspace{2mm} \Delta T_\textsf{water}$  ---- We can now solve for the *specific heat capacity* of the metal. <br/> $\large c_\textsf{metal} \hspace{2mm} = - \displaystyle\frac{ m_\textsf{water} \hspace{2mm} c_\textsf{water} \hspace{2mm} \Delta T_\textsf{water} } { m_\textsf{metal} \hspace{2mm} \Delta T_\textsf{metal} }$  ---- Plugging in our values... <br/> $$ \large c_\textsf{metal} \hspace{2mm} = - \displaystyle\frac{ \left( 100 \, \textsf{g} \right) \left( 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}} \right) \left( 2.6 ^\circ \textsf{C} \right) } { \left( 70.0 \, \textsf{g} \right) \left( -55.4 ^\circ \textsf{C} \right) } $$  $$ = 0.281 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}} $$  ---- The heat loss by the metal is <br/> $$ q_\textsf{metal} \hspace{2mm} = m_\textsf{metal} \hspace{2mm} c_\textsf{metal} \hspace{2mm} \Delta T_\textsf{metal} \hspace{2mm} $$  <br/> $$ = \left( 70.0 \, \textsf{g} \right) \left( 0.281 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}} \right) \left( -55.4 ^\circ \textsf{C} \right) $$  $$ = -1090 \, \textsf{J} $$  ---- This is the same as $- q_\textsf{water}$ <br/> $$ -q_\textsf{water} \hspace{2mm} = -m_\textsf{water} \hspace{2mm} c_\textsf{water} \hspace{2mm} \Delta T_\textsf{water} \hspace{2mm} $$  <br/> $$ = \left( 100 \, \textsf{g} \right) \left( 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}} \right) \left( 2.6 ^\circ \textsf{C} \right) $$  $$ = -1090 \, \textsf{J} $$  ---- :::info :point_right: Note that $-q_\textsf{water} \hspace{2mm}$ could have been calculated initially and used as a convenient intermediate in the calculations. ::: --- ## Question 23 <p style="text-align:left">If a reaction produces 1.506 kJ of heat, which is trapped in 30.0 g of water initially at 26.5 °C in a calorimeter like that in <a href="https://openstax.org/books/chemistry-2e/pages/5-2-calorimetry#CNX_Chem_05_02_Calorim">Figure 5.12</a>, what is the resulting temperature of the water?</p> ---- First, we gather our variables. <br/> $q_\textsf{rxn} = -1.506 \, \textsf{kJ} = -1506 \, \textsf{J}$  $m_\textsf{w} = 30.0 \, \textsf{g}$  $c_\textsf{w} = 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  $T^i_\textsf{w} = 26.5 \, ^\circ\textsf{C}$  ---- The heat released by the reaction was absorbed by the water. $q_\textsf{w} = -q_\textsf{rxn} = 1506 \, \textsf{J}$  <br/> Let's solve for change in temperature.  $q_\textsf{w} = m_\textsf{w} c_\textsf{w} \Delta T_\textsf{w}$  <br/> $\Delta T_\textsf{w} = \displaystyle\frac{q_\textsf{w}}{m_\textsf{w}c_\textsf{w}} = \displaystyle\frac{1506 \, \textsf{J}}{(30.0 \, \textsf{g})(4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}})} = 12.0 \, ^\circ\textsf{C}$  ---- Now we can solve for the final temperature. <br/> $\Delta T_\textsf{w} = T_f - T^i_\textsf{w}$  $T_f = \Delta T_\textsf{w} + T^i_\textsf{w} = 12.0 \, \textsf{C} + 26.5 \, \textsf{C} = 38.5 \, \textsf{C}$  --- ## Question 24 <p style="text-align:left">A 0.500-g sample of KCl is added to 50.0 g of water in a calorimeter (Figure 5.12). If the temperature decreases by 1.05 °C, what is the approximate amount of heat involved in the dissolution of the KCl, assuming the specific heat of the resulting solution is 4.18 J/g °C? Is the reaction exothermic or endothermic?</p> ---- Gather our variables <br/> $m_{\ce{KCl}} = 0.500 \, \textsf{g}$  $m_\textsf{water} = 50.0 \, \textsf{g}$  $\Delta T = -1.05 ^\circ \textsf{C}$  $c = 4.18 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  ---- The total mass of solution is $m = m_{\ce{KCl}} + m_\textsf{water} = 0.500 \, \textsf{g} + 50.0 \, \textsf{g} = 50.5 \, \textsf{g}$  <br/> The heat released by the water is  $q = mc \Delta T = \left( 50.5 \, \textsf{g} \right) \left( 4.18 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}} \right) \left( -1.05 ^\circ \textsf{C} \right) = -222 \, \textsf{J}$  <br/> This heat was absorbed by the dissolution of the KCl  $q_\textsf{dissolution} \hspace{3mm} = -q = 222 \, \textsf{J}$ ==Endothermic==  --- ## Question 26 <p style="text-align:left">When 50.0 g of 0.200 M NaCl(aq) at 24.1 °C is added to 100.0 g of 0.100 M AgNO<sub>3</sub>(aq) at 24.1 °C in a calorimeter, the temperature increases to 25.2 °C as AgCl(s) forms. Assuming the specific heat of the solution and products is 4.20 J/g °C, calculate the approximate amount of heat in joules produced.</p> ---- Gather variables <br/> $\large m_{\ce{NaCl}} \normalsize = 50.0 \, \textsf{g}$  $\large m_{\ce{AgNO3}} \normalsize = 100.0 \, \textsf{g}$  $T_i = 24.1 ^\circ \textsf{C}$  $T_f = 25.2 ^\circ \textsf{C}$  $c = 4.20 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  ---- The total mass of rxn mixture is $\large m = m_{\ce{NaCl}} \, + m_{\ce{AgNO3}} \, = \normalsize 50.0 \, \textsf{g} + 100.0 \, \textsf{g} = 150.0 \, \textsf{g}$  <br/> The change in temperature is  $\Delta T = T_f - T_i = 25.2 ^\circ \textsf{C} - 24.1 ^\circ \textsf{C} = 1.1 ^\circ \textsf{C}$  ---- Finally, we can calculate the heat produced by the reaction <br/> $$ q_\textsf{rxn} = -q_\textsf{soln} = mc \Delta T $$  <br/> $$ = - \left( 150.0 \, \textsf{g} \right) \left( 4.20 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}} \right) \left( 1.1 ^\circ \textsf{C} \right) $$  $$ = 690 \, \textsf{g} $$  --- ## Heat Capacity of Calorimeter **In a perfect world, a calorimeter would not absorb any heat. But in the real world we need to account for the heat absorbed by the calorimeter.** $$ q_{\textsf{Process}} \hspace{2mm} + q_{\textsf{Surrounding}} \hspace{3mm} + q_{\textsf{Calorimeter}} \hspace{3mm} = 0 $$ $$ q_{\textsf{Calorimeter}} \hspace{3mm} = - q_{\textsf{Process}} \hspace{2mm} - q_{\textsf{Surrounding}} $$ $$ c_p \Delta T = - q_{\textsf{Process}} \hspace{2mm} - q_{\textsf{Surrounding}} $$ $$ \large c_p = - \left[ \frac{q_{\textsf{Process}} + q_{\textsf{Surrounding}}}{\Delta T} \right] $$ --- This allows us to apply a correction to the calorimetry equations. $$ q_\textsf{rxn} + q_\textsf{surr} + q_\textsf{cal} = 0 $$ $$ q_\textsf{rxn} = - q_\textsf{surr} - q_\textsf{cal} $$ $$ m_\textsf{rxn} \, c_\textsf{rxn} \, \Delta T_\textsf{rxn} \, = - m_\textsf{surr} \, c_\textsf{surr} \, \Delta T_\textsf{surr} - C_p \Delta T_\textsf{surr} $$ $$ m_\textsf{rxn} \, c_\textsf{rxn} \, \left( \Delta T^f_\textsf{rxn} \, - \, \Delta T^i_\textsf{rxn} \right) = - \left( m_\textsf{surr} \, c_\textsf{surr} + C_p \right) \, \left( \Delta T^f_\textsf{surr} \, - \, \Delta T^i_\textsf{surr} \right) $$ :::info :point_right: Note that this assumes the calorimeter comes to thermal equilibrium with its contents. ::: --- ## Question 31 <p style="text-align:left">When a 0.740-g sample of trinitrotoluene (TNT), C<sub>7</sub>H<sub>5</sub>N<sub>2</sub>O<sub>6</sub>, is burned in a bomb calorimeter, the temperature increases from 23.4 °C to 26.9 °C. The heat capacity of the calorimeter is 534 J/°C, and it contains 675 mL of water. How much heat was produced by the combustion of the TNT sample?</p> ---- First, we gather our variables. $V_{\textsf{w}} = 675 \, \textsf{mL}$  $m_{\textsf{TNT}} = 0.740 \, \textsf{g}$  $T_i = 23.4 \, ^\circ\textsf{C}$  $T_f = 26.9 \, ^\circ\textsf{C}$  $\Delta T = T_f - T_i = 26.9 \, ^\circ\textsf{C} - 23.4 \, ^\circ\textsf{C} = 3.5 \, ^\circ\textsf{C}$  $c_p = 534 \, \frac{\textsf{J}}{^\circ\textsf{C}}$  $c = 4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}}$  ---- Heat enters the calorimeter($q_\textsf{c}$), enters the water ($q_\textsf{w}$) and leaves the reaction ($q_\textsf{rxn}$). The sum total of heats is zero. <br/> $\large q_{\textsf{c}} + q_{\textsf{w}} + q_{\textsf{rxn}} = 0$ ---- The heat absorbed by the calorimeter is $q_{\textsf{c}} = c_p\Delta T = \left( 534 \, \frac{\textsf{J}}{^\circ\textsf{C}} \right) (3.5 \, ^\circ\textsf{C}) = 1870 \, \textsf{J}$  <br/> The heat absorbed by the water is  $q_{\textsf{w}} = mc \Delta T = V\rho c \Delta T$  $= (675 \, \textsf{mL})\left(\frac{1 \, \textsf{g}}{1 \, \textsf{mL}}\right)(4.184 \, \frac{\textsf{J}}{\textsf{g}\cdot^\circ\textsf{C}})(3.5 \, ^\circ\textsf{C}) = 9880 \, \textsf{J}$  ---- Finally, we can solve for the heat released by the reaction. <br/> $\large q_{\textsf{rxn}} = -q_{\textsf{c}} - q_{\textsf{w}} = -1870 \, \textsf{J} - 9880 \, \textsf{J} = -11800 \, \textsf{J}$  --- ## Question 37 <p style="text-align:left">A serving of a breakfast cereal contains 3 g of protein, 18 g of carbohydrates, and 6 g of fat. What is the Calorie content of a serving of this cereal if the average number of Calories for fat is 9.1 Calories/g, for carbohydrates is 4.1 Calories/g, and for protein is 4.1 Calories/g?</p> ---- <p style="text-align:left">The energy from each individual component is</p> $(3 \, \textsf{g protein})\left(\displaystyle\frac{4.1 \, \textsf{Cal}}{1 \, \textsf{g}}\right) = 12.3 \, \textsf{Cal}$  $(18 \, \textsf{g carb})\left(\displaystyle\frac{4.1 \, \textsf{Cal}}{1 \, \textsf{g}}\right) = 73.8 \, \textsf{Cal}$  $(6 \, \textsf{g fat})\left(\displaystyle\frac{9.1 \, \textsf{Cal}}{1 \, \textsf{g}}\right) = 54.6 \, \textsf{Cal}$  <p style="text-align:left">Now we sum them all together.</p>  $12.3 \, \textsf{Cal} + 73.8 \, \textsf{Cal} + 54.6 \, \textsf{Cal} = 140.7 \, \textsf{Cal}$