3.1: PASS Thermochemistry: calculate q for cooling a compound (3.E.14)
- Page ID
- 464541
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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 x 103 g of ethylene glycol is cooled from 115.0°C to 85.0°C.
- Answer
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q = -265 kJ
Refer to LibreText 3.5, Calorimetry.
- Strategy Map
-
Step Hint 1. Identify what information you are given in the question. specific heat capacity
initial and final temperature
mass of ethylene glycol
2. Identify what the question is asking you to solve for. q when ethylene glycol is cooled 3. Recall an equation that connects the provided information and what you are looking for. You are given the specific heat capacity. It has the units J/gk. Use these units to help recall your calorimetry equation.
- Solution
-
\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 \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}) \\
& \mathrm{q}=-264990 \mathrm{~J} \\
& \mathrm{q}=-264.990 \mathrm{~kJ} \\
& \mathrm{q}=-265 \mathrm{~kJ}
\end{aligned}
- Guided Solution
-
Download Guided Solution as a pdf
Guided Solution
Hint
This 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.
Refer to: LibreText 3.5, Calorimetry
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 x 103 g of ethylene glycol is cooled from 115.0°C to 85.0°C. Provided Information:
Specific heat capacity (cp) = 2.42 J/gk
Mass m= 3.65x103g
Temperature Change (∆T)=Final T−Initial T
∆T= Tf−Ti = 85.0°C − 115.0°C = 30.0°C = 30.0 K
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? We are looking for q, 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. You are given the specific heat capacity. It has the units J/gk use these units to help recall your calorimetry equation.
Specific heat capacity (cp) = 2.42 𝐉/𝐠𝐤
J=Joules=unit of energy=q
g=grams=unit of mass=m
K=Kelvin=unit of temperature= ∆T
\begin{gathered}c_p=\frac{q}{m \Delta T}\end{gathered}
Then rearrange for q,
q=cpm∆T
Complete Solution Recall the calorimetry equation that uses specific heat capacity:
\mathrm{q}=\mathrm{c}_{\mathrm{p}} \mathrm{m} \Delta \mathrm{T}
Use the provided information to plug into the equation.\begin{gathered}
\left(\mathrm{c}_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}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
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.
Why does this answer make chemical sense?
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 in this situation the system 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.
Make sure you calculate ∆T correctly.
∆T= Tf − Ti
(question source from page titled 3.E: Thermochemistry (Exercises), 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), shared under a CC BY-NC-SA 3.0 license, authored, remixed, and/or curated by LibreTexts.