Calculate Initial Temperature T1: Physics Problem Solved

Hey guys! Let's dive into a classic physics problem involving heat transfer. We're given some masses, temperatures, and the specific heat capacity of water, and our mission is to find the initial temperature, T1. This kind of problem pops up all the time in thermodynamics, and understanding how to solve it can really boost your physics skills. We'll break it down step by step, so it's super clear and easy to follow. Ready to get started?

Problem Statement

So, here’s the scenario: We have two masses. Mass 1 (m1) is 3 kg, and mass 2 (m2) is 4 kg. We know that the final temperature of mass 2 (T2) is 80 degrees Celsius, and the overall final temperature (T) when the two masses reach thermal equilibrium is 45 degrees Celsius. We're also given the specific heat capacity of water, C(H2O), which is 4180 J/kg•K. Our main goal here is to figure out the initial temperature (T1) of mass 1. This involves understanding the principles of heat exchange and applying the relevant formulas. Let's explore how we can approach this problem and what concepts we need to consider.

To really grasp what's happening, imagine you've got two containers of water at different temperatures. When you mix them, they'll exchange heat until they reach the same temperature. That final temperature depends on the masses of the water and their initial temperatures. In our problem, we’re working backward – we know the final temperature and need to find one of the initial temperatures. The key to solving this is the principle of conservation of energy, which states that the heat lost by the hotter object equals the heat gained by the colder object, assuming no heat is lost to the surroundings. This is a crucial concept in thermodynamics and is the foundation for solving many similar problems. By understanding this, you're not just plugging numbers into a formula, but you're understanding the underlying physics.

Key Concepts to Remember

Before we jump into the calculations, let's quickly recap some essential concepts. The first key concept is thermal equilibrium. This is the state where two objects in contact have reached the same temperature, and there's no more net heat transfer between them. Think of it like two friends finally agreeing on a movie to watch – they've reached a consensus! In our problem, the final temperature (T) of 45 degrees Celsius represents this equilibrium state. Secondly, specific heat capacity (C) is a big player here. It tells us how much heat energy is needed to raise the temperature of 1 kg of a substance by 1 degree Celsius (or Kelvin). Water has a relatively high specific heat capacity, meaning it takes a lot of energy to change its temperature. This is why it’s used in many cooling systems! Understanding the role of specific heat capacity helps us appreciate why different materials heat up or cool down at different rates. Finally, the heat transfer equation is our main tool: Q = mcΔT, where Q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. This equation is like the secret sauce for solving heat-related problems, so make sure you're comfortable with it. With these concepts in mind, we're well-equipped to tackle the calculations!

Setting Up the Equations

Okay, let's get down to the nitty-gritty of setting up the equations. This is where we translate the word problem into mathematical expressions. Remember the principle of conservation of energy we talked about earlier? It's our guiding light here. The heat lost by mass 2 (Q2) will be equal to the heat gained by mass 1 (Q1). Mathematically, we can write this as Q1 = -Q2. The negative sign is crucial because it indicates that heat is being lost from mass 2, hence the temperature decrease, and gained by mass 1, leading to a temperature increase.

Now, let's use the heat transfer equation, Q = mcΔT, to express Q1 and Q2 in terms of the given variables. For mass 1, the heat gained (Q1) can be written as: Q1 = m1 * C(H2O) * (T - T1). Here, m1 is the mass of the first object, C(H2O) is the specific heat capacity of water, T is the final equilibrium temperature, and T1 is the initial temperature we're trying to find. Notice that the change in temperature (ΔT) is calculated as the final temperature minus the initial temperature (T - T1). For mass 2, the heat lost (Q2) is: Q2 = m2 * C(H2O) * (T - T2). Similarly, m2 is the mass of the second object, and T2 is its initial temperature. It’s important to keep track of which values are given and which we're solving for. Recognizing the knowns and unknowns will help us plug the correct numbers into the equation later on.

By substituting these expressions back into our conservation of energy equation, we get: m1 * C(H2O) * (T - T1) = - [m2 * C(H2O) * (T - T2)]. This equation is the heart of our problem. It connects all the given information and allows us to solve for T1. Notice that the specific heat capacity of water, C(H2O), appears on both sides of the equation. This is a common occurrence in these types of problems, and it often means we can simplify the equation by canceling out this term. Simplifying equations is a valuable skill in physics, as it reduces the complexity and the chances of making mistakes. In the next step, we'll do just that and then rearrange the equation to isolate T1.

Solving for T1

Alright, let's roll up our sleeves and get to solving for T1! We've got our main equation: m1 * C(H2O) * (T - T1) = - [m2 * C(H2O) * (T - T2)]. The first thing we can do to simplify this equation is to notice that C(H2O) is present on both sides. Since it's a common factor, we can divide both sides by C(H2O), effectively canceling it out. This simplifies our equation to: m1 * (T - T1) = - m2 * (T - T2). See how much cleaner that looks? Simplifying like this not only makes the calculations easier but also reduces the chances of making mistakes along the way. It's a pro tip for any physics problem!

Now, let's plug in the values we have: m1 = 3 kg, m2 = 4 kg, T = 45 degrees Celsius, and T2 = 80 degrees Celsius. Substituting these values, we get: 3 * (45 - T1) = - 4 * (45 - 80). Next, we'll expand both sides of the equation. On the left side, we have: 3 * 45 - 3 * T1 = 135 - 3T1. On the right side, we first calculate the term inside the parentheses: 45 - 80 = -35. Then, multiply by -4: -4 * (-35) = 140. So our equation now looks like this: 135 - 3T1 = 140. We're getting closer to isolating T1!

To isolate T1, we need to rearrange the equation. Let's subtract 135 from both sides: -3T1 = 140 - 135, which simplifies to -3T1 = 5. Finally, to solve for T1, we'll divide both sides by -3: T1 = 5 / (-3). So, T1 = -5/3, which is approximately -1.67 degrees Celsius. But hold on! A negative temperature might seem a bit odd in this context. Let's double-check our calculations and see if this makes sense in the real world. Sometimes, a negative temperature could indicate that we might have made a sign error or misinterpreted a part of the problem. So, let’s take a moment to reflect and ensure our solution is physically plausible.

Interpretation and Conclusion

Okay, we've crunched the numbers, and we've arrived at an initial temperature (T1) of approximately -1.67 degrees Celsius. Now comes the critical part: interpreting our result and making sure it makes sense in the real world. This is where physics isn't just about math; it's about understanding the physical implications of our answers. Let's think about what this negative temperature means in the context of our problem.

We started with mass 2 at 80 degrees Celsius, and the final equilibrium temperature was 45 degrees Celsius. This tells us that heat flowed from mass 2 to mass 1. If mass 1 started at a negative temperature, it would indeed gain heat and warm up to the final temperature. So, at first glance, the negative temperature doesn't immediately scream “error.” However, it's always good to pause and reflect on whether the magnitude of the temperature change and the final result are reasonable. In our scenario, we have a relatively small negative temperature, which means mass 1 started only slightly below the freezing point of water. This is plausible, especially if we're considering an environment where temperatures can drop below freezing.

So, after carefully reviewing our calculations and considering the physical context, we can confidently conclude that the initial temperature T1 is approximately -1.67 degrees Celsius. This problem illustrates a common scenario in thermodynamics where objects at different temperatures exchange heat until they reach thermal equilibrium. We've successfully applied the principle of conservation of energy and the heat transfer equation to solve for an unknown initial temperature. Remember, guys, the key to these problems is not just plugging in numbers but understanding the underlying concepts and making sure your answers make sense! Keep practicing, and you'll become a pro at solving these kinds of physics problems.