Heat Capacity & Specific Heat Calculation: UFPR Physics Problem
Hey guys! Today, we're diving into a classic physics problem from UFPR (Universidade Federal do Paraná) that involves calculating heat capacity and specific heat. This is a super important topic, especially if you're gearing up for the ENEM (Exame Nacional do Ensino Médio) or any other physics-heavy exam. We'll break down the problem step-by-step, so you can totally nail similar questions in the future. Let's get started!
Understanding the Problem: A Deep Dive
Okay, so let's break down the problem. We've got 500 grams of some mysterious substance that we need to heat up. The initial temperature is 20 °C, and we want to raise it to 70 °C. Now, here's the kicker: it takes 4000 calories to make this happen. The question is, what are the heat capacity and specific heat of this substance? This is where our knowledge of thermodynamics comes into play. Heat capacity and specific heat are fundamental concepts in understanding how materials respond to changes in temperature. You really need to grasp these concepts, because they pop up everywhere in physics and chemistry. We're not just solving a problem here; we're building a solid foundation.
Defining Heat Capacity
First off, let's talk about heat capacity. Heat capacity (C) is basically how much heat energy you need to add to an object to raise its temperature by 1 degree Celsius (or 1 Kelvin, since the temperature difference is the same). Think of it like this: a substance with a high heat capacity can absorb a lot of heat without a huge temperature change, while a substance with a low heat capacity heats up (or cools down) more quickly. It's a property of the entire object or system, not just the material it's made of. So, a giant block of aluminum will have a higher heat capacity than a tiny aluminum cube, even though they're the same material. We often express heat capacity in units of calories per degree Celsius (cal/°C) or Joules per Kelvin (J/K).
Specific Heat Explained
Now, let's move on to specific heat. Specific heat (c) is a more intrinsic property of a substance. It tells you how much heat energy it takes to raise the temperature of 1 gram of the substance by 1 degree Celsius. So, specific heat is like a fingerprint for a material – it helps us identify what the substance is. Water, for example, has a relatively high specific heat, which is why it's so good at regulating temperature. Metals, on the other hand, usually have lower specific heats. The units for specific heat are typically calories per gram per degree Celsius (cal/g°C) or Joules per kilogram per Kelvin (J/kg·K). The key difference to remember is that heat capacity is for a specific amount of a substance, while specific heat is per unit mass.
The Key Formula
To tackle this UFPR problem, we need to remember the fundamental equation that links heat, mass, specific heat, and temperature change. This equation is: Q = mcΔT, where:
- Q is the heat energy transferred (in calories or Joules)
- m is the mass of the substance (in grams or kilograms)
- c is the specific heat of the substance (in cal/g°C or J/kg·K)
- ΔT is the change in temperature (in °C or K)
This equation is your best friend for solving these types of problems. It's the bread and butter of calorimetry, which is the science of measuring heat. Mastering this formula is crucial, guys, seriously. In our case, we know Q (4000 calories), m (500 grams), and the initial and final temperatures, so we can calculate ΔT and then solve for c. Once we have specific heat, finding heat capacity is a breeze.
Solving the UFPR Problem Step-by-Step
Alright, let's get our hands dirty and actually solve this UFPR physics problem. We'll break it down into simple steps so it's super clear. We're not just looking for the answer; we're aiming to understand the process so you can tackle similar problems with confidence. Stick with me, and you'll see how straightforward this can be!
Step 1: Calculate the Temperature Change (ΔT)
The first thing we need to figure out is how much the temperature changed. This is super easy. We just subtract the initial temperature from the final temperature. So, in our case:
ΔT = Final Temperature - Initial Temperature ΔT = 70 °C - 20 °C ΔT = 50 °C
See? Simple as pie! We know the temperature increased by 50 degrees Celsius. This ΔT value is crucial because it tells us the magnitude of the temperature change that the substance experienced. It's a key ingredient in our main equation. Miscalculating this can throw off your entire solution, so always double-check your subtraction. This step is all about laying the groundwork for the rest of the calculation. A solid start makes the rest of the journey much smoother!
Step 2: Calculate the Specific Heat (c)
Now, we're going to use that famous formula we talked about earlier: Q = mcΔT. Remember, we know Q (4000 calories), m (500 grams), and we just calculated ΔT (50 °C). Our mission is to find 'c', the specific heat. So, we need to rearrange the formula to solve for 'c':
c = Q / (mΔT)
Now, we just plug in the values:
c = 4000 calories / (500 grams * 50 °C) c = 4000 calories / 25000 g°C c = 0.16 cal/g°C
Boom! We've found the specific heat. It's 0.16 calories per gram per degree Celsius. This value tells us something really important about the substance. It means that it takes 0.16 calories of heat energy to raise the temperature of 1 gram of this substance by 1 degree Celsius. This is a characteristic property of the material, and it helps us understand how it interacts with heat. We're not just crunching numbers here; we're uncovering the thermal behavior of a substance. Feels pretty cool, right?
Step 3: Calculate the Heat Capacity (C)
Okay, we've got specific heat in the bag. Now, let's find heat capacity. Remember, heat capacity is the amount of heat needed to raise the entire object's temperature by 1 degree Celsius. There are a couple of ways we can do this. The easiest is to use the relationship between heat capacity (C) and specific heat (c): C = mc
Where:
- C is the heat capacity
- m is the mass
- c is the specific heat
We already know m (500 grams) and c (0.16 cal/g°C), so let's plug them in:
C = 500 grams * 0.16 cal/g°C C = 80 cal/°C
There you have it! The heat capacity of our 500-gram substance is 80 calories per degree Celsius. This means that it takes 80 calories of heat energy to raise the temperature of the entire 500-gram sample by 1 degree Celsius. Notice how this is different from specific heat, which is per gram. Heat capacity is a property of the entire object, while specific heat is a property of the material itself. We're building a complete picture of how this substance responds to heat. It's all about understanding the connections between these concepts.
Step 4: Putting it All Together
So, we've calculated both the specific heat and the heat capacity. We found that the specific heat (c) is 0.16 cal/g°C and the heat capacity (C) is 80 cal/°C. Now, let's look back at the original question. It asks for both values, and we've got them! We've successfully navigated the problem and found the key thermal properties of this mysterious substance. We started with the fundamental equation Q = mcΔT, and we've used it to unlock the answers. This is the power of understanding the underlying principles. When you grasp the concepts, you can apply them to a wide range of problems. We're not just memorizing formulas; we're building problem-solving skills that will serve you well in physics and beyond. High five for that!
Choosing the Correct Answer and Wrapping Up
Okay, so we've done the heavy lifting and crunched the numbers. Now, let's circle back to the original question format, which presented multiple-choice options. In this case, the correct answer would be the one that matches our calculations: 80 cal/°C for heat capacity and 0.16 cal/g°C for specific heat.
Why This Matters: Real-World Applications
But hey, it's not just about getting the right answer on a test. It's about understanding why this stuff matters in the real world. Heat capacity and specific heat are crucial in tons of applications. Think about cooking: pots and pans are often made of materials with high heat capacity so they can distribute heat evenly. Or consider climate: the high specific heat of water helps regulate temperatures on Earth. And in engineering, these concepts are vital for designing everything from engines to cooling systems. So, by mastering these concepts, you're not just acing physics exams; you're gaining insights into the world around you. That's pretty awesome!
Final Thoughts and Tips for Success
So, guys, that's how you tackle a problem involving heat capacity and specific heat. Remember the key formula (Q = mcΔT), understand the difference between heat capacity and specific heat, and break the problem down into manageable steps. Don't be afraid to rearrange formulas and plug in values. Physics is all about applying the right tools to the job. And most importantly, practice, practice, practice! The more problems you solve, the more comfortable you'll become with these concepts. You've got this!
I hope this breakdown was helpful. If you have any questions or want to see more examples, let me know in the comments. Keep studying hard, and I'll catch you in the next one!