Calculating The Final Temperature Of Water Mixture A Physics Problem

In the realm of thermodynamics, understanding how different masses of substances at varying temperatures interact when mixed is a fundamental concept. This article delves into a specific scenario: determining the final temperature when a quantity of water at one temperature is mixed with another quantity of water at a different temperature. This problem utilizes the principle of heat exchange, where the heat lost by the hotter substance is equal to the heat gained by the colder substance, assuming no heat is lost to the surroundings. This concept is crucial in various applications, from engineering to everyday life, such as in heating and cooling systems, cooking, and even weather patterns. By meticulously analyzing the given parameters and applying the relevant formulas, we can accurately predict the equilibrium temperature of the mixture.

We are given a classic physics problem involving heat transfer and thermal equilibrium. Specifically, we have 4.0 kg of water initially at a temperature of 50°C. This warmer water is then mixed thoroughly with 6.0 kg of water that is initially at a cooler temperature of 15°C. The objective is to determine the final temperature of the mixture once thermal equilibrium is reached. To solve this, we will employ the principle of heat exchange, which states that the heat lost by the warmer water will be equal to the heat gained by the cooler water, assuming the system is closed and no heat is lost to the surroundings. We are also given the specific heat capacity of water, which is a crucial parameter in calculating the heat transfer, as 4200 J kg⁻¹°C⁻¹.

Before diving into the solution, let's clarify the fundamental concepts and formulas that underpin this problem. The core principle at play here is the law of conservation of energy, specifically as it applies to heat transfer. When two substances at different temperatures come into contact, heat will flow from the hotter substance to the colder one until they reach the same temperature, a state known as thermal equilibrium. The amount of heat transferred depends on several factors, including the mass of the substances, their specific heat capacities, and the temperature difference. The key formula we'll use to quantify heat transfer is:

Q = mcΔT

Where:

• Q represents the amount of heat transferred (in Joules)
• m represents the mass of the substance (in kilograms)
• c represents the specific heat capacity of the substance (in Joules per kilogram per degree Celsius)
• ΔT represents the change in temperature (in degrees Celsius)

The specific heat capacity (c) is a crucial property of a substance, representing the amount of heat energy required to raise the temperature of 1 kilogram of the substance by 1 degree Celsius. Water has a relatively high specific heat capacity, which means it can absorb a significant amount of heat without a drastic change in temperature. This property is why water is an excellent coolant and plays a vital role in regulating Earth's climate. In this problem, since we are dealing with water as both substances, the specific heat capacity will be the same for both. However, in scenarios involving different substances, it's crucial to consider their respective specific heat capacities. Another key concept is the principle of heat exchange, which can be mathematically expressed as:

Heat lost by hotter substance = Heat gained by colder substance

This principle is a direct application of the law of conservation of energy. In our problem, the heat lost by the warmer water will be equal to the heat gained by the cooler water. By setting up this equation using the Q = mcΔT formula for both the heat lost and heat gained, we can solve for the unknown final temperature of the mixture. It is essential to remember that ΔT is always calculated as the final temperature minus the initial temperature. For the hotter substance, ΔT will be negative (since it's losing heat), while for the colder substance, ΔT will be positive (since it's gaining heat). By carefully applying these concepts and formulas, we can accurately determine the final temperature of the water mixture.

Step-by-Step Solution

Now, let's walk through the solution step-by-step, applying the concepts and formulas discussed earlier. This will involve setting up the equations, plugging in the given values, and solving for the unknown final temperature.

1. Identify Given Variables

First, let's clearly list the information provided in the problem statement:

• Mass of warmer water (m₁) = 4.0 kg
• Initial temperature of warmer water (T₁₁) = 50°C
• Mass of cooler water (m₂) = 6.0 kg
• Initial temperature of cooler water (T₂₁) = 15°C
• Specific heat capacity of water (c) = 4200 J kg⁻¹°C⁻¹
• We need to find the final temperature of the mixture (Tf), which will be the same for both water quantities once they reach thermal equilibrium.

2. Apply the Heat Exchange Principle

As discussed earlier, the heat lost by the warmer water is equal to the heat gained by the cooler water. We can express this mathematically as:

Heat lost = Heat gained

Using the formula Q = mcΔT, we can expand this equation:

m₁ * c * (T₁₁ - Tf) = m₂ * c * (Tf - T₂₁)

Notice that the change in temperature (ΔT) is expressed differently for the warmer and cooler water. For the warmer water, it's the initial temperature minus the final temperature (T₁₁ - Tf), reflecting the temperature decrease. For the cooler water, it's the final temperature minus the initial temperature (Tf - T₂₁), reflecting the temperature increase.

3. Substitute Values into the Equation

Now, let's substitute the given values into the equation:

4. 0 kg * 4200 J kg⁻¹°C⁻¹ * (50°C - Tf) = 6.0 kg * 4200 J kg⁻¹°C⁻¹ * (Tf - 15°C)

4. Simplify the Equation

We can simplify the equation by canceling out the specific heat capacity (c) since it's the same for both sides and non-zero:

4. 0 kg * (50°C - Tf) = 6.0 kg * (Tf - 15°C)

Now, distribute the masses:

200 kg°C - 4.0 kg * Tf = 6.0 kg * Tf - 90 kg°C

5. Solve for Tf (Final Temperature)

To solve for Tf, let's rearrange the equation to group the terms with Tf on one side and the constant terms on the other side:

200 kg°C + 90 kg°C = 6.0 kg * Tf + 4.0 kg * Tf

Combine the terms:

290 kg°C = 10.0 kg * Tf

Now, divide both sides by 10.0 kg to isolate Tf:

Tf = 290 kg°C / 10.0 kg

Tf = 29°C

Therefore, the final temperature of the mixture is 29°C. This means that when 4.0 kg of water at 50°C is mixed with 6.0 kg of water at 15°C, the resulting mixture will reach a thermal equilibrium at 29°C.

Conclusion

In conclusion, by applying the principles of heat exchange and the formula Q = mcΔT, we successfully determined the final temperature of the water mixture to be 29°C. This problem highlights the practical application of thermodynamic principles in understanding how heat transfer occurs between substances at different temperatures. The step-by-step solution demonstrated the importance of carefully identifying given variables, applying the correct formulas, and meticulously solving the resulting equations. Understanding these concepts is crucial not only in physics but also in various engineering disciplines and everyday applications where temperature regulation is essential. From designing efficient heating and cooling systems to understanding weather patterns, the principles of heat transfer play a vital role. Furthermore, this problem illustrates the concept of thermal equilibrium, a fundamental concept in thermodynamics, where systems tend to reach a state where the temperature is uniform throughout. By grasping these principles, we can better understand and predict the behavior of systems involving heat transfer and temperature changes.

The final temperature of the mixture is 29°C.