Calorimeter Heat Absorption Calculation: A Chemistry Problem

Hey guys! Ever wondered how we measure the heat absorbed or released in a chemical reaction? Well, that's where calorimeters come in handy! Today, we're diving into a classic chemistry problem that involves calculating the heat absorbed by a calorimeter. Let's break it down step by step and make sure we all understand how to tackle this kind of question. Grab your calculators, and let's get started!

Understanding the Calorimetry Basics

Before we jump into the calculation, let's quickly recap the basic principles of calorimetry. Calorimetry is the science of measuring the heat associated with chemical reactions or physical changes. A calorimeter is the device used for this purpose. The basic idea is that when a reaction occurs inside the calorimeter, it either releases heat (exothermic reaction) or absorbs heat (endothermic reaction). This heat exchange causes a change in the temperature of the calorimeter and its contents. By measuring this temperature change, we can determine the amount of heat involved in the reaction.

The key formula we use in calorimetry is q = mC_pΔT, where:

• q represents the heat absorbed or released (in Joules).
• m is the mass of the substance (in grams).
• C_p is the specific heat capacity of the substance (in J/(g•°C)).
• ΔT is the change in temperature (in °C).

Specific heat capacity (C_p) is a crucial concept here. It tells us how much heat is required to raise the temperature of 1 gram of a substance by 1 degree Celsius. Different materials have different specific heat capacities. For example, water has a relatively high specific heat capacity (4.184 J/(g•°C)), which means it takes a lot of heat to change its temperature. Metals, on the other hand, generally have lower specific heat capacities.

Breaking Down the Problem

Now, let's look at the specific problem we're trying to solve. We're given the following information:

• The temperature decreases to 27.45°C. This tells us that the calorimeter has cooled down, meaning it has absorbed heat. The reaction inside is likely endothermic, pulling heat from the calorimeter.
• The calorimeter has a mass (m) of 1.400 kg. We'll need to convert this to grams since our specific heat is in J/(g•°C). So, 1.400 kg = 1400 g.
• The calorimeter has a specific heat (C_p) of 3.52 J/(g•°C). This is the amount of heat required to raise the temperature of 1 gram of the calorimeter by 1 degree Celsius.

We're asked to find the heat absorbed (q) by the reaction. To do this, we'll use the formula q = mC_pΔT. But wait, we're missing one crucial piece of information: ΔT, the change in temperature. The problem only tells us the final temperature (27.45°C), but not the initial temperature. Let's assume there's an initial temperature implicitly stated or missing from the original problem. For the sake of demonstration, let's suppose the initial temperature was 28.00°C. This will allow us to calculate the temperature change and proceed with the problem.

Step-by-Step Calculation

Okay, guys, let's get our hands dirty with some calculations! We're going to use the formula q = mC_pΔT to find out how much heat was absorbed. Remember, each part of this formula is super important, so let's take it slow and steady.

Step 1: Calculate the Change in Temperature (ΔT)

First things first, we need to figure out how much the temperature changed. Remember, ΔT (delta T) means the change in temperature, and we calculate it by subtracting the initial temperature from the final temperature:

ΔT = Final Temperature - Initial Temperature

We've assumed our initial temperature was 28.00°C and the final temperature is given as 27.45°C. So,

ΔT = 27.45°C - 28.00°C = -0.55°C

Notice the negative sign! This is super important because it tells us the temperature decreased. The calorimeter lost heat, which means the reaction absorbed it (endothermic, remember?). Keep that negative sign handy!

Step 2: Plug the Values into the Formula

Now for the fun part: plugging our values into the main formula:

q = mC_pΔT

We know:

• m (mass) = 1400 g (converted from 1.400 kg)
• C_p (specific heat capacity) = 3.52 J/(g•°C)
• ΔT (change in temperature) = -0.55°C

Let's plug 'em in:

q = (1400 g) * (3.52 J/(g•°C)) * (-0.55°C)

Step 3: Do the Math!

Time to crunch those numbers. Grab your calculators, folks! Let's multiply it all together:

q = 1400 * 3.52 * -0.55

q = -2712.16 J

Whoa, that's a number! But what does it mean? Remember that negative sign? It indicates that heat was absorbed by the reaction. The calorimeter lost this heat, which is why the temperature went down.

Step 4: Consider Significant Figures and Rounding

In chemistry, we always need to think about significant figures. Looking back at our original values:

• 1. 400 kg has four significant figures (1400 g also has four)
• 2. 52 J/(g•°C) has three significant figures
• -0.55°C has two significant figures (this limits our answer)

So, our final answer should be rounded to two significant figures. That means:

q ≈ -2700 J

Step 5: Convert to Kilojoules (Optional)

Sometimes, we prefer to express heat in kilojoules (kJ) instead of Joules (J). To do this, we simply divide by 1000:

q ≈ -2700 J / 1000 = -2.7 kJ

So, we can say that approximately 2.7 kJ of heat was absorbed by the reaction.

Analyzing the Answer and the Given Options

Alright, let's recap what we've found. We calculated that the heat absorbed by the reaction is approximately -2700 J (or -2.7 kJ). Remember, the negative sign just tells us the direction of heat flow – in this case, heat was absorbed by the reaction from the calorimeter.

Now, let's think about the options provided in the original problem:

• A. 140 J
• B. 418 J
• C. 1,470 J

Based on our calculation (-2700 J), none of these options seem to match directly. However, this is where we need to consider the possibility of an implicit or missing initial temperature in the original problem statement. Our calculation was based on the assumed initial temperature of 28.00°C. If the actual initial temperature was different, the ΔT would change, and consequently, the heat absorbed (q) would also change.

The Importance of Initial Conditions

This highlights a crucial point in calorimetry problems: initial conditions matter! Without knowing the initial temperature accurately, we can't determine the precise amount of heat absorbed or released. This is why it's so important to have complete information when setting up a calorimetry experiment.

What if the Initial Temperature Was Different?

Let's explore how a different initial temperature would affect our result. Suppose, for instance, the initial temperature was closer to the final temperature, say 27.50°C. Then:

ΔT = 27.45°C - 27.50°C = -0.05°C

Now, if we plug this new ΔT into our formula:

q = (1400 g) * (3.52 J/(g•°C)) * (-0.05°C)

q = -246.4 J

Rounding to two significant figures, we get:

q ≈ -250 J

See how much the answer changes? This illustrates the sensitivity of heat calculations to temperature changes. If the actual initial temperature was such that the heat absorbed was closer to one of the given options (140 J, 418 J, or 1,470 J), we would have selected that option. However, based on our initial assumption, our calculated answer is significantly larger than the provided options.

Final Thoughts and Key Takeaways

So, what have we learned today, guys? We've walked through a calorimetry problem step by step, calculating the heat absorbed by a reaction using the formula q = mC_pΔT. We've seen the importance of understanding the concepts of specific heat capacity and temperature change. We've also highlighted the critical role of initial conditions in calorimetry experiments.

The key takeaways from this problem are:

• Calorimetry is used to measure heat changes in reactions.
• The formula q = mC_pΔT is fundamental in calorimetry calculations.
• ΔT (change in temperature) is calculated as Final Temperature - Initial Temperature.
• Specific heat capacity (C_p) is a material property that indicates how much heat is required to change its temperature.
• Significant figures are crucial in reporting accurate results.
• Initial conditions, especially initial temperature, significantly impact the calculated heat.

Remember, guys, chemistry is all about practice and understanding the underlying principles. So, keep practicing these types of problems, and you'll become calorimetry pros in no time! If you have any questions, don't hesitate to ask. Keep exploring, keep learning, and I'll catch you in the next one! Happy calculating! 🚀🧪