Calculating Water Bath Temperature After Heat Transfer

by THE IDEN 55 views

In this article, we will delve into a fascinating chemistry problem involving heat transfer between a chemical reaction and its surroundings. Specifically, we'll explore a scenario where a reaction occurs inside a flask immersed in a water bath. The reaction releases heat, which then flows into the water bath, causing its temperature to rise. Our goal is to calculate the final temperature of the water bath after this heat transfer occurs. This type of calculation is fundamental in calorimetry, a branch of chemistry concerned with measuring the heat of chemical reactions or physical changes. Understanding these principles is crucial in various fields, including chemical engineering, materials science, and even cooking, where precise temperature control is essential. Let’s dive into the specifics of the problem and learn how to solve it step by step.

Problem Statement

We are presented with a scenario where a chemical reaction takes place inside a flask that is submerged in a water bath. The water bath initially contains 4.90 kg of water, and its initial temperature is 21.4 °C. During the reaction, 149 kJ of heat flows out of the flask and into the water bath. The task is to calculate the new temperature of the water in the bath after this heat transfer. This problem is a classic example of a calorimetry problem, where we are concerned with the measurement of heat flow. To solve this, we'll need to use the principles of thermodynamics, specifically the concept of heat capacity and the relationship between heat transfer, mass, specific heat, and temperature change. This problem provides a practical application of these concepts, demonstrating how they can be used to predict the temperature change in a system undergoing heat exchange. The careful measurement and calculation of heat transfer are essential in many scientific and industrial processes, making this a valuable skill to develop.

Key Concepts and Formulas

To solve this problem effectively, we need to understand a few key concepts and formulas related to heat transfer and calorimetry. First and foremost is the concept of heat capacity, which is the amount of heat required to raise the temperature of a substance by one degree Celsius (or one Kelvin). A related term is specific heat capacity, often denoted as c, which is the heat capacity per unit mass of the substance. Water has a relatively high specific heat capacity, approximately 4.184 J/g°C, meaning it takes a significant amount of heat to change its temperature compared to many other substances. This property makes water an excellent medium for water baths, as it can absorb a considerable amount of heat without undergoing drastic temperature changes.

The formula that connects heat transfer (q), mass (m), specific heat capacity (c), and temperature change (ΔT) is:

q = m c ΔT

Where:

  • q is the heat transferred (in Joules)
  • m is the mass of the substance (in grams)
  • c is the specific heat capacity of the substance (in J/g°C)
  • ΔT is the change in temperature (in °C), which is the final temperature (Tfinal) minus the initial temperature (Tinitial)

ΔT = Tfinal - Tinitial

This formula is the cornerstone of calorimetry calculations. It allows us to quantify the amount of heat gained or lost by a substance based on its mass, specific heat capacity, and the change in its temperature. In our problem, we know the heat transferred to the water bath, the mass of the water, and the specific heat capacity of water. We also know the initial temperature of the water. Using this information, we can rearrange the formula to solve for the final temperature of the water bath. Understanding and applying this formula is crucial for solving a wide range of calorimetry problems.

Step-by-Step Solution

Now, let's apply the concepts and formula discussed above to solve the problem step by step.

  1. Identify the knowns:

    • Mass of water (m) = 4.90 kg = 4900 g (We need to convert kg to g for consistency with the specific heat capacity units)
    • Initial temperature of water (Tinitial) = 21.4 °C
    • Heat transferred (q) = 149 kJ = 149,000 J (We need to convert kJ to J for consistency with the specific heat capacity units)
    • Specific heat capacity of water (c) = 4.184 J/g°C

  2. Rearrange the formula to solve for the change in temperature (ΔT):

    We start with the formula: q = m c ΔT

    Dividing both sides by (m c), we get:

    ΔT = q / (m c)

  3. Plug in the known values and calculate ΔT:

    ΔT = 149,000 J / (4900 g * 4.184 J/g°C)

    ΔT ≈ 7.29 °C

  4. Calculate the final temperature (Tfinal):

    We know that ΔT = Tfinal - Tinitial

    Rearranging to solve for Tfinal, we get:

    Tfinal = ΔT + Tinitial

    Tfinal = 7.29 °C + 21.4 °C

    Tfinal ≈ 28.69 °C

Therefore, the new temperature of the water bath after the heat transfer is approximately 28.69 °C. This step-by-step solution demonstrates the application of the heat transfer formula and the importance of unit conversions in calorimetry problems. By carefully identifying the knowns, rearranging the formula, and performing the calculations, we can accurately determine the final temperature of the water bath.

Detailed Calculation and Explanation

To further clarify the solution, let's delve into a more detailed calculation and explanation of each step. This will not only reinforce the concepts but also highlight the importance of attention to detail in scientific calculations. We'll revisit each step, providing additional insights and rationale behind the calculations.

Step 1: Identifying the Knowns and Converting Units

As previously mentioned, the first step in solving any problem is to identify the knowns. This involves extracting the relevant information from the problem statement and organizing it in a clear and concise manner. In this case, we have the mass of the water, the initial temperature, the amount of heat transferred, and the specific heat capacity of water. However, it's crucial to ensure that all the values are in consistent units before proceeding with the calculations. The mass of water was given in kilograms (kg), but the specific heat capacity of water is typically expressed in Joules per gram per degree Celsius (J/g°C). Therefore, we converted the mass from kilograms to grams by multiplying by 1000: 4.90 kg * 1000 g/kg = 4900 g. Similarly, the heat transferred was given in kilojoules (kJ), but we need it in Joules (J) to match the units of specific heat capacity. We converted kilojoules to Joules by multiplying by 1000: 149 kJ * 1000 J/kJ = 149,000 J. These unit conversions are essential to ensure that the final answer is in the correct units and that the calculations are accurate.

Step 2: Rearranging the Formula

The core formula for this problem is q = m c ΔT. This formula relates the heat transferred (q) to the mass (m), specific heat capacity (c), and the change in temperature (ΔT). However, our goal is to find the final temperature of the water, so we need to rearrange the formula to solve for ΔT. To do this, we divide both sides of the equation by (m c), which isolates ΔT on one side of the equation. This algebraic manipulation is a fundamental skill in problem-solving, allowing us to express the variable we want to find in terms of the known variables.

Step 3: Plugging in the Values and Calculating ΔT

Once we have the formula rearranged to solve for ΔT, we can plug in the known values. We substitute the values we identified in Step 1 into the formula: ΔT = 149,000 J / (4900 g * 4.184 J/g°C). Now, we perform the calculation. First, we multiply the mass and specific heat capacity: 4900 g * 4.184 J/g°C ≈ 20,501.6 J/°C. Then, we divide the heat transferred by this result: 149,000 J / 20,501.6 J/°C ≈ 7.29 °C. This result, 7.29 °C, represents the change in temperature of the water bath. It's important to pay attention to significant figures during these calculations. The final answer for ΔT should be rounded to the appropriate number of significant figures based on the given values in the problem.

Step 4: Calculating the Final Temperature

Finally, to find the final temperature of the water bath, we use the relationship ΔT = Tfinal - Tinitial. We know ΔT and Tinitial, so we can rearrange the formula to solve for Tfinal. Adding Tinitial to both sides of the equation, we get: Tfinal = ΔT + Tinitial. Plugging in the values, we have: Tfinal = 7.29 °C + 21.4 °C ≈ 28.69 °C. This is the final temperature of the water bath after the heat transfer. We can round this result to an appropriate number of significant figures based on the given values. In this case, the initial temperature has three significant figures, and the heat transferred also has three significant figures, so we can round the final temperature to three significant figures, giving us 28.7 °C. This detailed calculation and explanation provide a comprehensive understanding of the steps involved in solving the problem, highlighting the importance of unit conversions, algebraic manipulation, and attention to detail in scientific calculations.

Common Mistakes to Avoid

When solving calorimetry problems, there are several common mistakes that students often make. Being aware of these pitfalls can help you avoid them and ensure accurate results. One of the most frequent errors is forgetting to convert units. As we saw in the step-by-step solution, it's crucial to ensure that all values are in consistent units before plugging them into the formula. For instance, if the specific heat capacity is in J/g°C, the mass must be in grams, and the heat transferred must be in Joules. Failing to make these conversions can lead to errors in the magnitude of the final answer.

Another common mistake is incorrectly rearranging the formula. The formula q = m c ΔT can be rearranged to solve for any of the variables, but it's essential to do so correctly. A simple algebraic error can lead to an incorrect result. It's always a good practice to double-check the rearranged formula before plugging in the values. For example, if you are solving for ΔT, make sure you have ΔT = q / (m c) and not something else.

A third common mistake is not paying attention to the sign of the heat transfer. Heat transfer (q) is positive when heat is added to the system (endothermic process) and negative when heat is released by the system (exothermic process). In our problem, heat flows out of the flask and into the water bath, so the heat transferred to the water bath is positive. However, if we were considering the heat change of the reaction inside the flask, the heat transferred would be negative. Ignoring the sign can lead to an incorrect calculation of the final temperature.

Finally, not considering significant figures is another common mistake. The final answer should be reported with the appropriate number of significant figures based on the given values. For example, if the initial temperature is given to three significant figures and the heat transferred is given to two significant figures, the final temperature should be reported to two significant figures. Neglecting significant figures can give a misleading impression of the precision of the result. By being mindful of these common mistakes and taking the time to double-check your work, you can improve your accuracy in solving calorimetry problems.

Practical Applications of Calorimetry

Calorimetry, the science of measuring heat, has a wide array of practical applications across various fields. Understanding these applications can highlight the importance of mastering the concepts and calculations involved in calorimetry. In the realm of chemistry, calorimetry is used extensively to determine the heat of reactions, which is a crucial piece of information for understanding the energy changes that occur during chemical processes. This information is vital for designing chemical reactions and processes, as it helps predict whether a reaction will release or absorb heat, and how much energy is involved. For instance, in the development of new drugs or materials, calorimetric measurements can help determine the stability and reactivity of the compounds.

In the food industry, calorimetry plays a significant role in determining the nutritional content of food. The caloric value of food, which is the amount of energy it provides when consumed, is measured using a bomb calorimeter. This device measures the heat released when a food sample is burned completely. This information is essential for nutrition labeling and dietary planning. Understanding the caloric content of different foods helps individuals make informed choices about their diet and energy intake.

Engineering also benefits significantly from calorimetry. In mechanical engineering, calorimetry is used to measure the heat generated by engines and other machinery, which is crucial for designing efficient cooling systems and preventing overheating. In chemical engineering, calorimetry is used to optimize chemical processes by understanding the heat flow involved in various reactions. This helps in designing reactors and other equipment that can handle the heat generated or absorbed by the reactions. Additionally, in materials science, calorimetry is used to study the thermal properties of materials, such as their specific heat capacity and thermal conductivity, which are important for selecting materials for different applications.

In environmental science, calorimetry is used to study the energy content of various fuels, including biofuels and fossil fuels. This helps in assessing the efficiency and environmental impact of different energy sources. Calorimetry is also used to study the metabolic rates of organisms, which is important for understanding ecological processes and the flow of energy in ecosystems. These diverse applications demonstrate the broad relevance of calorimetry in various scientific and industrial contexts.

Conclusion

In conclusion, calculating the final temperature of a water bath after heat transfer from a chemical reaction is a fundamental problem in calorimetry. By understanding the concepts of heat capacity and specific heat capacity, and by applying the formula q = m c ΔT, we can accurately determine the temperature change in the water bath. This problem highlights the importance of unit conversions, careful calculations, and attention to detail in scientific problem-solving. Moreover, we discussed common mistakes to avoid and the practical applications of calorimetry in various fields, demonstrating the broad relevance of this topic. Mastering these concepts and skills is crucial for anyone studying chemistry, physics, or related fields, as they provide a foundation for understanding energy transfer and thermal processes in a wide range of contexts. Whether you're designing chemical reactions, analyzing the nutritional content of food, or studying the thermal properties of materials, calorimetry provides the tools and knowledge to make accurate measurements and informed decisions.