Calculating Energy Released From Propane Combustion A Step-by-Step Guide

Hey guys! Today, we're diving into a cool chemistry problem: figuring out how much energy is released when we burn a sample of propane in a bomb calorimeter. It might sound intimidating, but trust me, we'll break it down step by step. Let's get started!

The Propane Combustion Problem

So, here's the scenario: We've got a propane sample ($C_3H_8$) with a mass of 0.47 g. We're burning this propane in a bomb calorimeter, which is basically a fancy, sealed container designed to measure the heat released during a reaction. This calorimeter has a mass of 1.350 kg and a specific heat of 5.82 J/(g⋅°C). The big question we want to answer is: How much energy is released during this combustion process?

To solve this, we'll need to use the principles of calorimetry, which is all about measuring heat transfer. The key idea here is that the heat released by the propane combustion is absorbed by the calorimeter. By measuring the temperature change of the calorimeter, we can calculate the amount of heat it absorbed, which is equal to the heat released by the reaction. Think of it like this: the propane is the source of the heat, and the calorimeter is the receiver. The amount of heat the receiver picks up tells us how much the source gave off.

Now, let's delve into the nitty-gritty details. We need to remember a crucial formula from thermodynamics: q = mcΔT. This simple yet powerful equation is the cornerstone of calorimetry. Let's break down what each part means:

• q: This represents the heat transferred, which is what we're trying to find. It's the amount of energy released by the propane combustion.
• m: This is the mass of the substance absorbing the heat, in our case, the calorimeter. It's super important to get the units right here! We're given the mass in kilograms, but the specific heat is in terms of grams, so we'll need to convert kilograms to grams.
• c: This is the specific heat capacity of the substance. It tells us how much energy it takes to raise the temperature of 1 gram of the substance by 1 degree Celsius. In our problem, this is the specific heat of the calorimeter.
• ΔT: This is the change in temperature, which is the difference between the final temperature and the initial temperature. The bigger the temperature change, the more heat was transferred.

With this formula in mind, we can start putting the pieces together. Remember, our goal is to find q, the heat released by the propane combustion. To do that, we'll need the mass of the calorimeter, its specific heat, and the change in temperature. We're already given the mass and specific heat, so the next step is to figure out how to use this information to calculate the heat. This is where the molar mass of propane comes into play, as it helps us relate the mass of propane burned to the amount of energy released. We also need to think about the enthalpy of combustion for propane, which tells us how much heat is released when one mole of propane is completely burned. By combining these concepts with our q = mcΔT equation, we'll be able to crack this problem and find the energy released. Stay tuned!

Step-by-Step Calculation of Energy Released

Okay, let's get into the actual calculations now! We know we need to use the formula q = mcΔT, but before we can plug in the numbers, we need to make sure we have all the information in the right units. Remember, the mass of the calorimeter is given in kilograms (1.350 kg), but the specific heat is in joules per gram per degree Celsius (5.82 J/(g⋅°C)). So, the first thing we need to do is convert the mass of the calorimeter from kilograms to grams. This is a pretty straightforward conversion: just multiply by 1000 (since there are 1000 grams in a kilogram).

So, the mass of the calorimeter in grams is 1.350 kg * 1000 g/kg = 1350 g. Awesome! Now we have the mass in the correct units. We also know the specific heat capacity (c) is 5.82 J/(g⋅°C). But, hold on a second! There's one crucial piece of information missing: the change in temperature (ΔT). The problem doesn't explicitly give us the change in temperature. What gives?

This is where we need to make a smart assumption and use the information we do have. We know the mass of propane burned (0.47 g) and we're looking for the total energy released. We also know that the energy released by burning the propane will be absorbed by the calorimeter, causing its temperature to rise. To relate the mass of propane to the energy released, we need to think about the enthalpy of combustion for propane. This is a value that tells us how much heat is released when one mole of propane is completely burned. You can usually find this value in a chemistry textbook or online resource.

The enthalpy of combustion for propane is approximately -2220 kJ/mol (the negative sign indicates that heat is released, which is an exothermic reaction). This means that when one mole of propane is burned, 2220 kilojoules of energy are released. But, we're not burning a whole mole of propane; we're burning 0.47 g. So, we need to figure out how many moles of propane that is.

To do this, we'll need the molar mass of propane ($C_3H_8$). We can calculate this by adding up the atomic masses of each element in the molecule: (3 * 12.01 g/mol for carbon) + (8 * 1.01 g/mol for hydrogen) = 44.11 g/mol. Now we can convert grams of propane to moles of propane:

0. 47 g propane * (1 mol propane / 44.11 g propane) = 0.01065 mol propane

So, we've burned approximately 0.01065 moles of propane. Now we can calculate the total heat released by the combustion:

1. 01065 mol propane * (-2220 kJ/mol) = -23.64 kJ

This means that the combustion of 0.47 g of propane releases 23.64 kJ of energy. But remember, the q in our q = mcΔT equation needs to be in joules, not kilojoules. So, let's convert that:

-2. 64 kJ * (1000 J/kJ) = -23640 J

Now we know the heat released (q) is -23640 J. We can plug this, along with the mass of the calorimeter and its specific heat, into our q = mcΔT equation and solve for ΔT:

-3. 640 J = (1350 g) * (5.82 J/(g⋅°C)) * ΔT

Now, let's isolate ΔT:

ΔT = -23640 J / (1350 g * 5.82 J/(g⋅°C))

ΔT = -23640 J / 7857 J/°C

ΔT = -3.01 °C

So, the temperature change of the calorimeter is -3.01 °C. But wait! A negative temperature change? That doesn't make sense in this context. We know the combustion released heat, so the temperature should have increased. This is a good reminder to always check if your answer makes sense in the context of the problem! The issue here is that we used the negative value for the heat released (-23640 J). While the negative sign is important to indicate that heat is being released (an exothermic reaction), for the q in the q = mcΔT equation, we're concerned with the magnitude of the heat transfer. So, we should use the absolute value of the heat released, which is 23640 J.

Let's recalculate ΔT using the positive value:

ΔT = 23640 J / (1350 g * 5.82 J/(g⋅°C))

ΔT = 23640 J / 7857 J/°C

ΔT = 3.01 °C

Okay, that makes much more sense! The temperature of the calorimeter increased by 3.01 °C. But, remember the original question: how much energy was released? We actually already calculated that! It's the 23640 J (or 23.64 kJ) we found earlier. The temperature change was just an intermediate step to help us verify our calculations and make sure everything was consistent.

So, the final answer is: The combustion of 0.47 g of propane releases approximately 23.64 kJ of energy. Woohoo! We did it!

Key Concepts and Takeaways

Alright, guys, let's recap the key concepts and takeaways from this propane combustion problem. We tackled a complex calorimetry problem, and hopefully, you've gained a solid understanding of the principles involved. Here's the lowdown:

1. Calorimetry is all about measuring heat transfer. We use calorimeters to measure the heat released or absorbed during chemical reactions or physical changes. The fundamental idea is that the heat lost by one system (like our propane combustion) is gained by another (the calorimeter).
2. The equation q = mcΔT is your best friend. This equation is the cornerstone of calorimetry. It relates the heat transferred (q) to the mass of the substance (m), its specific heat capacity (c), and the change in temperature (ΔT). Make sure you understand what each term means and how to use the equation correctly.
3. Units matter! Pay close attention to the units of each value. In our problem, we had to convert kilograms to grams to ensure consistency with the specific heat units. Always double-check your units and make conversions when necessary.
4. Enthalpy of combustion is key for relating mass to energy. The enthalpy of combustion tells you how much heat is released when one mole of a substance is completely burned. We used this value, along with the molar mass of propane, to calculate the heat released from our specific mass of propane.
5. Always check if your answer makes sense. In our calculation, we initially got a negative temperature change, which didn't make sense for a combustion reaction. This prompted us to re-examine our steps and catch a mistake. Always think critically about your results and see if they align with your expectations.
6. Breaking down the problem into smaller steps is crucial. Complex problems can seem overwhelming, but if you break them down into smaller, manageable steps, they become much easier to solve. We systematically worked through the problem, converting units, calculating moles, finding the heat released, and finally, calculating the temperature change. This step-by-step approach is a valuable problem-solving strategy in chemistry and beyond.

By understanding these key concepts and takeaways, you'll be well-equipped to tackle similar calorimetry problems in the future. Remember, practice makes perfect! So, try working through some more examples and don't be afraid to ask for help if you get stuck.

Applications of Calorimetry

So, we've spent a good amount of time discussing the theory and calculations behind calorimetry, but you might be wondering, "Okay, that's cool, but where is this actually used in the real world?" Great question! Calorimetry is not just a theoretical concept confined to textbooks and classrooms. It has a wide range of practical applications in various fields, from chemistry and materials science to food science and even medicine.

One of the most common applications of calorimetry is in determining the nutritional content of food. You've probably seen calorie information on food labels, right? Well, those calorie values are often determined using a technique called bomb calorimetry, which is exactly what we were working with in our propane combustion problem! Food scientists burn a sample of food in a calorimeter and measure the heat released. This heat energy is then converted into calories (or kilocalories) to provide the nutritional information we see on labels. This is crucial for understanding the energy content of different foods and for developing balanced diets.

In the field of chemistry, calorimetry is used extensively to study chemical reactions and their thermodynamics. Scientists use calorimeters to measure the heat of reaction (also known as enthalpy change) for various chemical processes. This information is vital for understanding reaction mechanisms, predicting reaction rates, and designing new chemical processes. For example, calorimetry can be used to determine the heat released or absorbed during the synthesis of a new drug, which is essential for scaling up the production process.

Materials science also benefits greatly from calorimetry. Researchers use calorimetry to study the thermal properties of materials, such as their specific heat capacity, melting point, and heat of fusion. This information is crucial for selecting appropriate materials for various applications, from building construction to aerospace engineering. For instance, calorimetry can be used to determine the heat capacity of a new alloy, which is important for predicting how it will respond to temperature changes in a particular application.

In the pharmaceutical industry, calorimetry plays a vital role in drug development and manufacturing. Calorimeters are used to study the thermal stability of drugs, determine their heat of solution, and optimize crystallization processes. This information is crucial for ensuring the quality, safety, and efficacy of pharmaceutical products. For example, calorimetry can be used to determine the optimal temperature for storing a vaccine to maintain its potency.

Even in medicine, calorimetry has some fascinating applications. For example, direct calorimetry can be used to measure the metabolic rate of a patient by measuring the heat they produce. This information can be helpful in diagnosing metabolic disorders and monitoring the energy expenditure of patients in intensive care. While direct calorimetry is not widely used due to its complexity, indirect calorimetry, which measures oxygen consumption and carbon dioxide production, is a more common technique for assessing metabolic rate.

These are just a few examples of the many applications of calorimetry. From determining the calories in your favorite snack to developing new drugs and materials, calorimetry is a powerful tool that helps us understand the world around us. So, the next time you see calorie information on a food label or hear about a new material with enhanced thermal properties, remember the principles of calorimetry and the important role they play in our daily lives.

Conclusion

So, there you have it, guys! We've successfully navigated the world of propane combustion and calorimetry. We started with a seemingly complex problem, but by breaking it down step-by-step, using the right equations, and paying attention to units, we were able to calculate the energy released during the combustion process. We also explored the broader applications of calorimetry in various fields, highlighting its importance in science and technology. Hopefully, this journey has not only equipped you with the skills to solve calorimetry problems but also sparked your curiosity about the fascinating world of thermodynamics and heat transfer. Keep exploring, keep learning, and remember that even the most challenging problems can be tackled with a systematic approach and a little bit of perseverance!