Thermochemistry: Calculate q for Cooling a Compound
Question
Ethylene glycol, used as a coolant in automotive engines, has a specific heat capacity of 2.42 J g-1 K-1. Calculate q for the system when 3.65 × 103 g of ethylene glycol is cooled from 115.0°C to 85.0°C.
Show/Hide Answer
q = -265 kJ
Refer to Section 3.5: Calorimetry (1).
Strategy Map
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1. Identify what information you are given in the question.
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2. Identify what the question is asking you to solve for.
Show/Hide Hintq when ethylene glycol is cooled |
3. Recall an equation that connects the provided information and what you are looking for.
Show/Hide HintYou are given the specific heat capacity. It has the units J/gk. Use these units to help recall your calorimetry equation. |
Solution
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[latex]\begin{aligned} &\mathrm{q}=\mathrm{c}_{\mathrm{p}}\mathrm{m}\Delta\mathrm{T}\\ &\mathrm{q}=\left(2.42\frac{\mathrm{J}}{\mathrm{gk}}\right)\left(3.65\times10^3\mathrm{~g}\right)\left(85.0^{\circ}\mathrm{C}-115.0^{\circ}\mathrm{C}\right)\\&\mathrm{q}=\left(2.42\frac{\mathrm{J}}{\mathrm{gk}}\right)\left(3.65\times10^3\mathrm{~g}\right)(-30.0\mathrm{~K})\\ &\mathrm{q}=-264990\mathrm{~J}\\&\mathrm{q}=-264.990\mathrm{~kJ}\\ &\mathrm{q}=-265\mathrm{~kJ} \end{aligned}[/latex]
Answer: q = -265 kJ
Guided Solution
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The guided solution below will give you the reasoning for each step to get your answer, with reminders and hints.
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This question is a calculation problem where your knowledge of calorimetry is tested by giving you experimental data to solve for the heat (q) lost by ethylene glycol when it cools.
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Ethylene glycol, used as a coolant in automotive engines, has a specific heat capacity of 2.42 J g-1 K-1. Calculate q when 3.65 × 103 g of ethylene glycol is cooled from 115.0°C to 85.0°C.
Show/Hide Think About This!Provided Information:
Looking for: q, heat lost by ethylene glycol when it cools |
If the ethylene glycol is cooled down, what can you expect about your final answer?
Show/Hide Think About This!We are looking for q, which is the heat gained or lost by the system. If the system is cooling down, it is losing energy in the form of heat. This means the final answer for q will be negative. |
Recall the different equations for calorimetry. Since we are using specific heat capacity, we will use the equation with cp.
Show/Hide Don’t Forget!You are given the specific heat capacity. It has the units J/gk use these units to help recall your calorimetry equation.
[latex]\begin{gathered} \mathrm{c}_\mathrm{p}=\dfrac{\mathrm{q}}{\mathrm{m}\Delta\mathrm{T}} \end{gathered}[/latex] Then rearrange for q, q = cpm∆T |
| Complete Solution |
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| Recall the calorimetry equation that uses specific heat capacity.
[latex]\begin{gathered} \mathrm{q}=\mathrm{c}_{\mathrm{p}}\mathrm{m}\Delta\mathrm{T} \end{gathered}[/latex] Use the provided information to plug into the equation. [latex]\begin{gathered} \left(\mathrm{c}_\mathrm{p}\right)=2.42\mathrm{~J}/\mathrm{gk}\\ (\mathrm{m})=3.65\times 10^3\mathrm{~g}\\ (\Delta\mathrm{T})=\mathrm{T}_{\mathrm{f}}-\mathrm{T}_{\mathrm{i}}=85.0^{\circ}\mathrm{C}-115.0^{\circ}\mathrm{C}=30.0^{\circ} \mathrm{C}\\ \mathrm{q}=\left(2.42\frac{\mathrm{J}}{\mathrm{gk}}\right)\left(3.65\times 10^3\mathrm{~g}\right)\left(85.0^{\circ}\mathrm{C}-115.0^{\circ}\mathrm{C}\right)\\ \mathrm{q}=\left(2.42\frac{\mathrm{J}}{\mathrm{gk}}\right)\left(3.65\times 10^3\mathrm{~g}\right)(-30.0\mathrm{~K}) \end{gathered}[/latex] Multiply out and solve for q: q = −264990 J Recall that the significant figures in your final answer should match the information given in the question and that q should be in kilojoules. q = −264.990 kJ Answer: q = −265 kJ |
Check Your Work
Since ethylene glycol (the system) is cooling down, it is losing energy in the form of heat. This means ΔT is negative, and we expect the final answer for q will be negative.
Show/Hide Watch Out!
Make sure you calculate ∆T correctly: ∆T = Tf − Ti
Does your answer make chemical sense?
Show/Hide Answer
Calorimetry is the process of calculating the energy change (or enthalpy change) during a chemical reaction. This is done by measuring the amount of heat lost or gained by the system.
With the data provided in the question, we know that the system in this situation was losing heat, and therefore, its energy change was negative. From this, we know the process was exothermic and released -265 kJ into its surroundings.
PASS Attribution
- LibreTexts PASS Chemistry Book CHEM 1510/1520 (2).
- Question 3.E.14 from LibreTexts PASS Chemistry Book CHEM 1510/1520 (3) is used under a CC BY-NC-SA 4.0 license.
- Question 3.E.14 is question 3.E.14 from LibreTexts TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520) (4), which is under a CC BY-NC-SA 3.0 license.
References
1. Thompson Rivers University. 3.5: Calorimetry. In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2022. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/03%3A_Thermochemistry/3.05%3A_Calorimetry.
2. Blackstock, L.; Brewer, S.; Jensen, A. In PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510%2F%2F1520.
3. Blackstock, L.; Brewer, S.; Jensen, A. 3.1: PASS Thermochemistry: Calculate q for Cooling a Compound (3.E.14). In PASS Chemistry Book CHEM 1510/1520; LibreTexts, 2024. https://chem.libretexts.org/Courses/Thompson_Rivers_University/PASS_Chemistry_Book_CHEM_1510_1520/03%3A_Thermochemistry/3.01%3A_3.X_Template
4. Thompson Rivers University. 3.E: Thermochemistry (Exercises). In TRU: Fundamentals and Principles of Chemistry (CHEM 1510 and CHEM 1520); LibreTexts, 2023. https://chem.libretexts.org/Courses/Thompson_Rivers_University/TRU%3A_Fundamentals_and_Principles_of_Chemistry_(CHEM_1510_and_CHEM_1520)/03%3A_Thermochemistry/3.E%3A_Thermochemistry_(Exercises).
