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

by Viktoria Ivanova 61 views

Hey guys! Today, we're diving into a classic chemistry problem: calculating the amount of energy released when a sample of propane (C3H8{C_3H_8}) is burned in a bomb calorimeter. This is a fundamental concept in thermochemistry, and understanding it helps us grasp how chemical reactions produce or consume energy. We'll break down the problem step-by-step, making it super easy to follow. Let's get started!

Okay, so here’s the scenario. We have a sample of propane that weighs 0.47 grams. This propane is burned inside a bomb calorimeter, which is basically a fancy container designed to measure heat changes accurately. The calorimeter itself has a mass of 1.350 kilograms, and its specific heat is 5.82 Joules per gram per degree Celsius J/(g⋅∘C){J/(g \cdot ^{\circ}C)}). Our mission, should we choose to accept it, is to figure out how much energy is released during this combustion process.

This problem is a great example of how we apply principles of calorimetry to real-world situations. Calorimetry is the science of measuring heat flow associated with chemical reactions or physical changes. A bomb calorimeter is a type of calorimeter specifically designed to measure the heat of combustion at constant volume. It's a closed system, meaning no mass can enter or leave, and it's built to withstand the high pressures that can develop during combustion reactions.

To solve this problem effectively, we'll need to use the principles of heat transfer and the properties of the calorimeter. We'll be relying on the equation: q=mcΔT{q = mc\Delta T} where:

  • q is the heat transferred (energy released or absorbed).
  • m is the mass of the substance.
  • c is the specific heat capacity of the substance.
  • \Delta T is the change in temperature.

By carefully applying this equation and considering the specifics of the problem, we can determine the energy released by the combustion of propane. So, let's roll up our sleeves and dive into the solution!

Step 1: Convert Mass to Grams

The first thing we need to do is make sure our units are consistent. The mass of the calorimeter is given in kilograms (kg), but the specific heat is given in Joules per gram per degree Celsius J/(g⋅∘C){J/(g \cdot ^{\circ}C)}). So, let’s convert the mass of the calorimeter from kilograms to grams. To do this, we multiply the mass in kilograms by 1000, since there are 1000 grams in a kilogram.

Mass of calorimeter in grams: 1.350 kg×1000gkg=1350 g{1.350 \text{ kg} \times 1000 \frac{\text{g}}{\text{kg}} = 1350 \text{ g}}

This conversion ensures that we're using the same units throughout our calculations, which is crucial for getting the correct answer. Remember, always double-check your units before plugging numbers into equations! It's a simple step that can save you from making mistakes.

Step 2: Calculate the Heat Capacity of the Calorimeter

The heat capacity of an object is the amount of heat required to raise its temperature by one degree Celsius. We can calculate the heat capacity (C{C}) of the calorimeter using the formula:

C=mc{C = mc}

where:

  • m is the mass of the calorimeter (in grams).
  • c is the specific heat of the calorimeter 5.82 J/(g⋅∘C){5.82 \text{ J/(g} \cdot ^{\circ}\text{C)}}).

Plugging in the values, we get:

C=1350 g×5.82Jg⋅∘C=7857J∘C{C = 1350 \text{ g} \times 5.82 \frac{\text{J}}{\text{g} \cdot ^{\circ}\text{C}} = 7857 \frac{\text{J}}{^{\circ}\text{C}}}

So, the heat capacity of the calorimeter is 7857 Joules per degree Celsius. This value tells us how much energy the calorimeter can absorb for each degree Celsius increase in temperature. This is a key piece of information that we’ll use in the next step to determine the total energy released by the combustion of propane.

Step 3: Determine the Temperature Change

To figure out the energy released, we need to know the temperature change (ΔT{\Delta T}) of the calorimeter. Unfortunately, the problem doesn’t directly give us the temperature change. This is a bit of a trick! We need to think about what the problem is asking and what information is essential.

However, since the problem asks how much energy is released, we can express the heat released in terms of the temperature change, leaving $\Delta T$ as a variable in our final expression.

Step 4: Calculate the Heat Released by Combustion

Now, let's calculate the heat (q{q}) released during the combustion of propane. We can use the formula:

q=CΔT{q = C\Delta T}

where:

  • C is the heat capacity of the calorimeter 7857 J/∘C{7857 \text{ J}/^{\circ}\text{C}}).
  • \Delta T is the change in temperature (which we'll keep as $\Delta T$ for now).

Plugging in the value for C, we get:

q=7857J∘C×ΔT{q = 7857 \frac{\text{J}}{^{\circ}\text{C}} \times \Delta T}

So, the heat released by the combustion is 7857ΔT{7857\Delta T} Joules. This expression tells us that the amount of heat released is directly proportional to the temperature change. The larger the temperature change, the more heat is released, and vice versa. This makes intuitive sense, right? A bigger temperature jump means more energy was given off by the reaction.

Step 5: Express the Final Answer

Finally, we can express the energy released during the combustion of 0.47 g of propane in terms of the temperature change, $\Delta T$. The heat released, q, is given by:

q=7857ΔT J{q = 7857 \Delta T \text{ J}}

This is our final answer! It tells us that for every degree Celsius the temperature of the calorimeter increases, 7857 Joules of energy are released. If we knew the actual temperature change, we could plug it into this equation to get a numerical value for the energy released.

And there you have it, folks! We’ve successfully calculated the energy released during the combustion of 0.47 grams of propane in a bomb calorimeter. We walked through each step, from converting units to calculating heat capacity and finally expressing the energy released in terms of the temperature change.

This problem showcases the power of calorimetry and how we can use it to measure the heat associated with chemical reactions. Understanding these concepts is super important in chemistry, as it helps us predict and control reactions, design new materials, and even develop energy technologies.

I hope this step-by-step breakdown has made the process clear and easy to understand. Remember, chemistry can be challenging, but with a little practice and a solid understanding of the fundamentals, you can tackle any problem that comes your way. Keep up the great work, and I’ll catch you in the next one!