Mixing Water Temperatures: Calculate Water Masses

by ADMIN 50 views

Hey guys! Today, we're diving into a classic physics problem: mixing water at different temperatures. Specifically, we'll tackle the question of how to mix water at 40°C and 70°C to get 40 kg of water at a final temperature of 70°C. It sounds like a kitchen conundrum, but it's a fantastic way to understand heat transfer and thermal equilibrium. So, grab your metaphorical beakers (or just your thinking caps!) and let's get started.

Understanding the Problem

First, let's break down what we're trying to achieve. We want to end up with 40 kg of water at 70°C. We have two sources: water that's already at 70°C and cooler water at 40°C. Our mission is to figure out how much of each we need to mix. This is where the magic of thermal equilibrium comes in. Thermal equilibrium is the state where two objects (in this case, our water masses) in contact reach the same temperature. Heat will naturally flow from the hotter object to the cooler one until they're both at the same temperature. So, the heat lost by the hotter water will be gained by the cooler water. To solve this, we'll be using the principle of conservation of energy, which, in this scenario, translates to the heat lost by the hot water equaling the heat gained by the cold water. We'll also need to use the formula for heat transfer, which relates the heat transferred to the mass, specific heat capacity, and the change in temperature. Don't worry, we'll walk through each step carefully!

Setting Up the Equations

To solve this problem, we'll need to set up a couple of equations. Let's define our variables:

  • m1: Mass of water at 70°C (in kg)
  • m2: Mass of water at 40°C (in kg)

We know that the total mass of the mixture should be 40 kg, so we can write our first equation:

m1 + m2 = 40

This is a simple equation, but it's crucial. It tells us that whatever amount of hot water we use, the remaining amount will be cold water to reach our 40 kg total. Now, let's think about the heat transfer. The heat lost by the hot water will be equal to the heat gained by the cold water. The formula for heat transfer is:

Q = m * c * ΔT

Where:

  • Q is the heat transferred (in Joules)
  • m is the mass (in kg)
  • c is the specific heat capacity of water (approximately 4186 J/kg°C)
  • ΔT is the change in temperature (in °C)

Since the specific heat capacity of water (c) is the same for both the hot and cold water, it will cancel out in our equation. The heat lost by the hot water is:

Q_lost = m1 * c * (70 - 70)

Notice that the final temperature is also 70°C. The heat gained by the cold water is:

Q_gained = m2 * c * (70 - 40)

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

m1 * c * (70 - 70) = m2 * c * (70 - 40)

Since c is the same on both sides, we can simplify this to:

m1 * (70 - 70) = m2 * (70 - 40)

But wait a minute! There seems to be an issue here. Since the final temperature is the same as the hot water's initial temperature (70°C), the term (70 - 70) becomes zero. This implies that the mass of the hot water (m1) doesn't affect the final temperature, which isn't quite right in this scenario. Let's revisit our understanding of the problem and see if we've missed anything.

Identifying the Misunderstanding

Okay, guys, let's take a step back. It seems we've hit a snag in our initial setup. The core issue is that the problem statement seems a bit contradictory. We're asked to obtain 40 kg of water at 70°C by mixing water at 70°C and 40°C. Logically, if we already have water at 70°C, adding more water at 70°C won't change the temperature! The final temperature will always be 70°C if we're only mixing water that's already at that temperature.

To make this problem solvable and meaningful, we need to adjust the final temperature we're aiming for. Let's assume the problem meant to ask: "How to mix water at 40°C and 70°C to obtain 40 kg of water at a different temperature, say 50°C?" This change makes the problem a classic heat transfer scenario. So, from here on out, we'll work with the assumption that the target temperature is 50°C. This highlights a crucial aspect of problem-solving: always double-check if the given information makes logical sense before diving into calculations!

Revisiting the Equations with the Corrected Final Temperature

Now that we've clarified the problem (aiming for a final temperature of 50°C), let's revisit our equations. Our first equation remains the same:

m1 + m2 = 40

This still represents the total mass of the mixture. However, our heat transfer equation needs to be updated to reflect the new final temperature. The heat lost by the hot water (initially at 70°C) is now:

Q_lost = m1 * c * (70 - 50)

The heat gained by the cold water (initially at 40°C) is:

Q_gained = m2 * c * (50 - 40)

Setting these equal to each other (and canceling out the c as before), we get:

m1 * (70 - 50) = m2 * (50 - 40)

Simplifying this gives us:

20 * m1 = 10 * m2

Which can be further simplified to:

2 * m1 = m2

This equation tells us that the mass of the cold water (m2) is twice the mass of the hot water (m1). Now we have a system of two equations:

  1. m1 + m2 = 40
  2. 2 * m1 = m2

Solving the System of Equations

We can use substitution to solve this system. Since we know that m2 = 2 * m1, we can substitute this into the first equation:

m1 + (2 * m1) = 40

Combining the terms with m1, we get:

3 * m1 = 40

Now, we can solve for m1:

m1 = 40 / 3

m1 ≈ 13.33 kg

So, we need approximately 13.33 kg of water at 70°C. Now we can find m2 using either of our original equations. Let's use m1 + m2 = 40:

13.  33 + m2 = 40

m2 = 40 - 13.33

m2 ≈ 26.67 kg

Therefore, we need approximately 26.67 kg of water at 40°C.

Verifying the Solution

It's always a good idea to check our answer to make sure it makes sense. We found that we need about 13.33 kg of water at 70°C and 26.67 kg of water at 40°C to get 40 kg of water at 50°C. Let's plug these values back into our heat transfer equation to see if they balance:

Q_lost = 13.33 * c * (70 - 50) = 13.33 * c * 20

Q_gained = 26.67 * c * (50 - 40) = 26.67 * c * 10

Q_lost ≈ 266.6 * c

Q_gained ≈ 266.7 * c

The values are very close (the slight difference is due to rounding), which confirms that our solution is correct! So, to get 40 kg of water at 50°C, you'd need to mix approximately 13.33 kg of water at 70°C with 26.67 kg of water at 40°C.

Key Takeaways and Practical Applications

This problem, even with the initial hiccup, highlights several important concepts in physics and problem-solving:

  • The principle of thermal equilibrium: Heat flows from hotter objects to cooler objects until they reach the same temperature.
  • Conservation of energy: In a closed system, energy is neither created nor destroyed, but it can be transferred from one form to another (in this case, heat transfer).
  • The heat transfer formula (Q = m * c * ΔT): This formula is fundamental for understanding how heat is transferred based on mass, specific heat capacity, and temperature change.
  • The importance of logical consistency: Always ensure the problem statement makes sense before diving into calculations. Identifying the initial contradiction in the problem was crucial for solving it correctly.
  • Solving systems of equations: Many physics problems involve multiple unknowns, requiring you to set up and solve a system of equations.

This type of calculation isn't just an academic exercise. It has practical applications in various fields, including:

  • Engineering: Designing heating and cooling systems, mixing chemicals in industrial processes, and ensuring proper temperatures in manufacturing.
  • Cooking: While we might not do these calculations exactly, understanding heat transfer helps us cook food evenly and achieve desired temperatures.
  • Meteorology: Predicting temperature changes in the atmosphere and understanding weather patterns.
  • Everyday life: Adjusting water temperatures for showers, baths, or making the perfect cup of tea!

Conclusion

So, there you have it! We've successfully tackled the problem of mixing water temperatures, learned some valuable physics principles, and even corrected a slightly flawed problem statement along the way. Remember, guys, physics is all about understanding the world around us, and even seemingly simple problems like this can offer deep insights into how things work. Keep those thinking caps on, and happy problem-solving!