Equilibrium Temperature Calculation: Calorimeter Problem

Hey guys! Let's dive into a classic physics problem: calculating the equilibrium temperature when we mix different substances in a calorimeter. This is a super important concept in thermodynamics, and understanding it will help you ace your physics exams and even understand how things work in the real world. In this article, we'll break down a specific problem step-by-step, making sure you grasp every detail. Our mission today is to figure out the final temperature when we pour hot water into a calorimeter. Ready to get started?

Understanding the Calorimeter and Heat Exchange

Before we jump into the calculations, let's make sure we understand what a calorimeter is and how heat exchange works. A calorimeter is basically an insulated container used to measure the heat involved in a chemical reaction or a physical change. Think of it as a super-insulated thermos that prevents heat from escaping or entering. This allows us to accurately measure the heat transfer between substances inside the calorimeter.

In our case, we have a calorimeter with a heat capacity of 200 Joules per kilogram Kelvin (J/kg·K). This value tells us how much heat energy is required to raise the temperature of 1 kg of the calorimeter by 1 Kelvin (which is the same as 1 degree Celsius). The calorimeter is initially at a temperature (t1) of 20 degrees Celsius. Now, we're going to pour in 200 grams of water at a temperature (t2) of 100 degrees Celsius. The hot water will transfer heat to the calorimeter, and eventually, they will both reach the same equilibrium temperature. Our goal is to calculate what that final temperature will be.

When we mix the hot water and the calorimeter, heat will flow from the hotter substance (the water) to the cooler substance (the calorimeter). This heat transfer will continue until both the water and the calorimeter reach the same temperature – the equilibrium temperature. The principle we'll use to solve this problem is the conservation of energy. This means that the heat lost by the hot water will be equal to the heat gained by the calorimeter. This is a fundamental concept in thermodynamics and is essential for understanding how heat transfer works. To solve this problem effectively, we'll use the formula for heat transfer, which relates the amount of heat transferred to the mass, specific heat capacity, and the change in temperature. We'll also need to make sure our units are consistent throughout the calculation. Are you ready to see how this works in action? Let's move on to setting up the problem.

Setting Up the Problem and Identifying Key Variables

Okay, let's get our ducks in a row and set up this problem so it's nice and clear. First, we need to identify all the key variables and their values. This will help us keep track of what we know and what we need to find out. Here’s what we’ve got:

• Calorimeter:
  • Heat capacity (C_cal): 200 J/kg·K
  • Initial temperature (t1): 20 °C
• Water:
  • Mass (m_water): 200 g (which we'll need to convert to kg)
  • Initial temperature (t2): 100 °C
• What we need to find:
  • Equilibrium temperature (T_eq)

Now, let's think about units. We have the mass of water in grams, but the heat capacity of the calorimeter is in terms of kilograms. To keep things consistent, we need to convert the mass of water from grams to kilograms. Remember, there are 1000 grams in a kilogram, so:

m_water = 200 g = 200 / 1000 kg = 0.2 kg

Great! We’ve got the mass of the water in kilograms. Next, we need to know the specific heat capacity of water. This is a constant value that tells us how much heat energy is required to raise the temperature of 1 kg of water by 1 degree Celsius. The specific heat capacity of water (c_water) is approximately 4200 J/kg·K. This is a crucial value for our calculation, so make sure you note it down.

Now that we have all the necessary variables, we can set up the equations we'll use to solve for the equilibrium temperature. We know that the heat lost by the water will be equal to the heat gained by the calorimeter. So, we can express this mathematically. This is a common approach in physics problems: identify the knowns, the unknowns, and the relationships between them. By setting up the problem in this way, we're making it much easier to solve. Are you ready to see how we can use these variables and the principle of conservation of energy to calculate the equilibrium temperature? Let's move on to the next step!

Applying the Heat Transfer Equation

Alright, now for the juicy part – applying the heat transfer equation! This is where we put our physics knowledge to work. As we discussed earlier, the heat lost by the water is equal to the heat gained by the calorimeter. We can express this mathematically using the following equation:

Heat lost by water = Heat gained by calorimeter

The heat transfer equation itself looks like this:

Q = mcΔT

Where:

• Q is the heat transferred (in Joules)
• m is the mass (in kg)
• c is the specific heat capacity (in J/kg·K)
• ΔT is the change in temperature (in °C or K)

For the water, the heat lost (Q_water) can be calculated as:

Q_water = m_water * c_water * (T_eq - t2)

Notice that the change in temperature (ΔT) is (T_eq - t2) because the water is cooling down from its initial temperature (t2) to the equilibrium temperature (T_eq). Since the water is losing heat, this value will be negative. For the calorimeter, the heat gained (Q_cal) can be calculated as:

Q_cal = C_cal * (T_eq - t1)

Here, we're using the heat capacity of the calorimeter (C_cal) directly, since it represents the heat required to change the temperature of the entire calorimeter, not just per unit mass. The change in temperature (ΔT) is (T_eq - t1) because the calorimeter is heating up from its initial temperature (t1) to the equilibrium temperature (T_eq). This value will be positive since the calorimeter is gaining heat.

Now, we can set the heat lost by the water equal to the heat gained by the calorimeter:

m_water * c_water * (T_eq - t2) = C_cal * (T_eq - t1)

This is our main equation! We have all the values we need except for T_eq, which is what we're trying to find. The next step is to plug in the values we identified earlier and solve for T_eq. This might look a bit intimidating, but don't worry, we'll break it down step by step. Are you ready to get those numbers plugged in and see how we can isolate T_eq?

Solving for the Equilibrium Temperature

Okay, let's get down to business and solve for the equilibrium temperature (T_eq). We have our equation from the previous step:

m_water * c_water * (T_eq - t2) = C_cal * (T_eq - t1)

Now, let's plug in the values we know:

0.  2 kg * 4200 J/kg·K * (T_eq - 100 °C) = 200 J/kg·K * (T_eq - 20 °C)

First, we can simplify the equation by multiplying the constants:

840 * (T_eq - 100) = 200 * (T_eq - 20)

Next, we distribute the numbers on both sides of the equation:

840 * T_eq - 84000 = 200 * T_eq - 4000

Now, we want to get all the terms with T_eq on one side and the constants on the other side. Let's subtract 200 * T_eq from both sides:

840 * T_eq - 200 * T_eq - 84000 = -4000

Which simplifies to:

640 * T_eq - 84000 = -4000

Now, let's add 84000 to both sides:

640 * T_eq = 80000

Finally, we divide both sides by 640 to isolate T_eq:

T_eq = 80000 / 640

T_eq = 125 °C

Wait a minute! This result seems a bit off. We started with water at 100 °C, and we’re getting an equilibrium temperature of 125 °C. That's impossible! What did we miss? Ah, it looks like we made a mistake in our sign convention. The heat lost by the water should be negative, and the heat gained by the calorimeter should be positive. Let's correct our equation:

- [m_water * c_water * (T_eq - t2)] = C_cal * (T_eq - t1)

Plugging in the values again:

- [0.2 kg * 4200 J/kg·K * (T_eq - 100 °C)] = 200 J/kg·K * (T_eq - 20 °C)

-840 * (T_eq - 100) = 200 * (T_eq - 20)

-840 * T_eq + 84000 = 200 * T_eq - 4000

Now, let's add 840 * T_eq to both sides and add 4000 to both sides:

84000 + 4000 = 200 * T_eq + 840 * T_eq

88000 = 1040 * T_eq

Divide both sides by 1040:

T_eq = 88000 / 1040

T_eq ≈ 84.62 °C

That makes much more sense! The equilibrium temperature is approximately 84.62 °C. This value is between the initial temperatures of the water and the calorimeter, which is what we would expect. Phew! It's always a good idea to double-check your work and make sure the answer makes logical sense. Now that we've successfully calculated the equilibrium temperature, let's summarize our findings and discuss the key takeaways.

Conclusion: Key Takeaways and Practical Implications

Woohoo! We did it! We successfully calculated the equilibrium temperature when 200 grams of water at 100 degrees Celsius is poured into a calorimeter with a heat capacity of 200 J/kg·K, initially at 20 degrees Celsius. Our final answer is approximately 84.62 °C.

Let's quickly recap the steps we took to solve this problem:

1. Understanding the Calorimeter and Heat Exchange: We defined what a calorimeter is and how it works to measure heat transfer. We also discussed the principle of conservation of energy, which states that the heat lost by one substance is equal to the heat gained by another in a closed system.
2. Setting Up the Problem and Identifying Key Variables: We listed all the known variables (heat capacity of the calorimeter, initial temperatures, mass of water) and the unknown variable (equilibrium temperature). We also converted the mass of water from grams to kilograms and identified the specific heat capacity of water.
3. Applying the Heat Transfer Equation: We used the formula Q = mcΔT to express the heat lost by the water and the heat gained by the calorimeter. We then set these two expressions equal to each other, remembering to account for the sign convention (heat lost is negative, heat gained is positive).
4. Solving for the Equilibrium Temperature: We plugged in the known values into the equation and solved for T_eq. We carefully performed the algebraic manipulations, double-checked our work, and corrected a sign error to arrive at the final answer.

This type of problem is not just an academic exercise. It has practical implications in many areas, such as:

• Chemistry: Calorimetry is used to measure the heat released or absorbed in chemical reactions. This information is crucial for understanding the thermodynamics of chemical processes.
• Engineering: Engineers use calorimetry principles to design heating and cooling systems, as well as to analyze the thermal performance of materials and devices.
• Food Science: Calorimetry is used to measure the caloric content of foods, which is important for nutrition and dietary planning.

Understanding how to calculate equilibrium temperature and apply the principles of heat transfer is a valuable skill in many fields. By breaking down the problem into manageable steps and carefully applying the relevant equations, we can solve complex thermal problems with confidence. So, next time you encounter a similar problem, remember the steps we followed, and you'll be well on your way to finding the solution. Keep practicing, and you'll become a heat transfer pro in no time! You got this!