Calculating Heat Release 500g Water Cools From 95.0C To 25.0C
Understanding heat transfer and calorimetry is crucial in various scientific and engineering applications. This article delves into a specific problem involving the cooling of water and the subsequent release of calories. We aim to determine the amount of heat energy, measured in calories, released when 500 grams of water cools from an initial temperature of 95.0°C to a final temperature of 25.0°C. This is a classic thermochemistry problem that utilizes the principle of specific heat capacity. The specific heat capacity of a substance is the amount of heat required to raise the temperature of 1 gram of the substance by 1 degree Celsius. For water, this value is approximately 1 calorie per gram per degree Celsius (1 cal/g°C). This means that it takes 1 calorie of heat to increase the temperature of 1 gram of water by 1°C. Conversely, when water cools, it releases 1 calorie of heat for every gram of water for every 1°C decrease in temperature. To solve this problem, we need to calculate the total temperature change experienced by the water and then use the specific heat capacity to determine the total heat released. This involves a straightforward application of the formula: Heat (Q) = mass (m) × specific heat capacity (c) × temperature change (ΔT). By understanding and applying this formula, we can accurately determine the caloric energy released during the cooling process. This concept is fundamental not only in chemistry but also in physics, engineering, and even everyday applications such as understanding how cooling systems work or how much energy is needed to heat water for various purposes.
Calculation of Heat Released
The calculation of heat released during the cooling process involves a straightforward application of the specific heat capacity formula. This formula, Q = m × c × ΔT, allows us to quantify the amount of heat energy transferred based on the mass of the substance, its specific heat capacity, and the change in temperature. In our scenario, we have 500 grams of water cooling from 95.0°C to 25.0°C. The first step in our calculation is to determine the temperature change (ΔT). This is found by subtracting the final temperature from the initial temperature: ΔT = Final Temperature - Initial Temperature. In this case, ΔT = 25.0°C - 95.0°C = -70.0°C. The negative sign indicates that heat is being released, as the water is cooling down. Next, we need the specific heat capacity (c) of water. As mentioned earlier, the specific heat capacity of water is approximately 1 calorie per gram per degree Celsius (1 cal/g°C). This value is a constant for water under normal conditions and is essential for our calculation. Now that we have all the necessary components, we can plug the values into the formula: Q = m × c × ΔT. Substituting the values, we get: Q = 500 g × 1 cal/g°C × (-70.0°C). Multiplying these values together, we find: Q = -35,000 calories. The negative sign confirms that this is the heat released by the water. Therefore, the amount of heat released when 500 grams of water cools from 95.0°C to 25.0°C is 35,000 calories. This calculation demonstrates the direct relationship between temperature change and heat transfer, a fundamental principle in thermodynamics. This principle is widely used in various applications, from designing cooling systems in engines to understanding climate patterns and weather phenomena.
Step-by-Step Solution
To provide a clear and concise understanding of the solution, let's break down the step-by-step process of calculating the heat released. This will not only clarify the calculation but also reinforce the underlying principles of thermochemistry. The problem states that 500 grams of water cools from an initial temperature of 95.0°C to a final temperature of 25.0°C. Our goal is to determine the amount of heat, measured in calories, that is released during this cooling process. Step 1: Identify the given values. We are given the mass of water (m) = 500 g, the initial temperature (Ti) = 95.0°C, and the final temperature (Tf) = 25.0°C. We also know the specific heat capacity of water (c) = 1 cal/g°C. Step 2: Calculate the temperature change (ΔT). The temperature change is calculated by subtracting the final temperature from the initial temperature: ΔT = Tf - Ti = 25.0°C - 95.0°C = -70.0°C. The negative sign indicates that the temperature has decreased, meaning heat is being released. Step 3: Apply the specific heat capacity formula. The formula to calculate the heat (Q) released or absorbed is: Q = m × c × ΔT. Step 4: Substitute the values into the formula. Plugging in the values we have: Q = 500 g × 1 cal/g°C × (-70.0°C). Step 5: Perform the calculation. Multiplying the values gives us: Q = -35,000 calories. The negative sign indicates that heat is released by the water as it cools. Step 6: Express the answer in scientific notation. The answer, -35,000 calories, can be expressed in scientific notation as -3.50 × 10^4 calories. This step is important for clarity and consistency, especially when dealing with large numbers. By following these steps, we have systematically calculated the heat released during the cooling of water. This process highlights the importance of understanding the relationships between mass, specific heat capacity, temperature change, and heat transfer. The principles used here are fundamental in many areas of science and engineering, making this a valuable skill to master.
Correct Answer and Explanation
The correct answer to the question, “How many calories are released when 500 g of water cools from 95.0°C to 25.0°C?” is E. 3.50 x 10^4 cal. This answer is derived from the calculations we performed earlier, where we found that 35,000 calories of heat are released when 500 grams of water cools from 95.0°C to 25.0°C. This value is equivalent to 3.50 x 10^4 calories when expressed in scientific notation. The explanation for this answer lies in the principles of thermochemistry and the application of the specific heat capacity formula. The formula, Q = m × c × ΔT, is the cornerstone of this calculation. Where: Q represents the heat released or absorbed, m is the mass of the substance, c is the specific heat capacity of the substance, and ΔT is the change in temperature. In our case, the mass (m) of the water is 500 g, the specific heat capacity (c) of water is 1 cal/g°C, and the change in temperature (ΔT) is -70.0°C (calculated as 25.0°C - 95.0°C). Plugging these values into the formula gives us: Q = 500 g × 1 cal/g°C × (-70.0°C) = -35,000 calories. The negative sign indicates that heat is released by the water as it cools. Converting this to scientific notation, we get 3.50 x 10^4 calories. The other options provided are incorrect because they do not accurately reflect the calculation using the specific heat capacity formula. Option A, 4.75 x 10^4 cal, is significantly higher than the correct value. Option B, 1.25 x 10^4 cal, is much lower. Options C and D, 70.0 cal and 35.0 cal respectively, are drastically smaller and do not account for the mass of the water or the magnitude of the temperature change. Therefore, understanding and correctly applying the specific heat capacity formula is essential for arriving at the correct answer. This problem underscores the importance of careful calculation and a solid grasp of the fundamental principles of thermodynamics.
Common Mistakes to Avoid
When solving thermochemistry problems, particularly those involving specific heat capacity, several common mistakes can lead to incorrect answers. Being aware of these pitfalls is crucial for achieving accurate results. One of the most frequent errors is incorrectly calculating the temperature change (ΔT). The temperature change is defined as the final temperature minus the initial temperature (ΔT = Tf - Ti). A common mistake is to subtract the temperatures in the wrong order, leading to a ΔT with the wrong sign. In our problem, the water cools from 95.0°C to 25.0°C, so ΔT should be 25.0°C - 95.0°C = -70.0°C. Forgetting the negative sign can lead to an incorrect interpretation of whether heat is being released or absorbed. Another common mistake is using the wrong units or forgetting to include units in the calculation. The specific heat capacity of water is 1 cal/g°C, so the mass must be in grams and the temperature in degrees Celsius for the units to cancel out correctly. If the mass were given in kilograms, it would need to be converted to grams before using the formula. Similarly, if the temperature were given in Fahrenheit, it would need to be converted to Celsius. A third mistake is using the incorrect specific heat capacity. While the specific heat capacity of water is approximately 1 cal/g°C, other substances have different specific heat capacities. Using the wrong value will result in an incorrect calculation. It’s essential to use the correct specific heat capacity for the substance in question. Another error arises from misunderstanding the formula Q = m × c × ΔT. This formula applies specifically to situations where there is a temperature change without a change in phase (e.g., solid to liquid or liquid to gas). If a phase change occurs, additional calculations involving latent heat are required. Finally, arithmetical errors in the calculation process can also lead to incorrect answers. It's important to double-check calculations and use a calculator when necessary to avoid mistakes. By being mindful of these common mistakes, students can improve their accuracy and confidence in solving thermochemistry problems.
Applications in Real Life
The principles of heat transfer and specific heat capacity, which we used to solve the problem of water cooling, have numerous real-life applications that impact our daily lives and various industries. Understanding how heat is absorbed or released during temperature changes is fundamental in many fields. One significant application is in heating and cooling systems. Whether it's the radiator in a car, the air conditioning in a building, or the central heating system in a home, all these systems rely on the principles of heat transfer. For instance, the radiator in a car uses a coolant (often water mixed with antifreeze) that absorbs heat from the engine. This heated coolant then circulates through the radiator, where the heat is dissipated into the air, cooling the engine. The efficiency of these systems depends on the specific heat capacity of the fluids used and the temperature differences involved. In the food industry, understanding heat transfer is crucial for cooking, pasteurization, and food preservation. Cooking involves transferring heat to food to change its texture and flavor. Pasteurization is a process that uses heat to kill harmful microorganisms in food and beverages, extending their shelf life. Similarly, food preservation techniques like canning and freezing rely on controlling heat transfer to prevent spoilage. Climate and weather patterns are also heavily influenced by heat transfer. The Earth's oceans and atmosphere act as massive heat reservoirs, absorbing and releasing heat to regulate global temperatures. Water's high specific heat capacity plays a crucial role in moderating coastal climates, as it takes more energy to change the temperature of water compared to land. This is why coastal areas tend to have milder temperatures than inland regions. In engineering, the principles of heat transfer are essential in designing engines, power plants, and electronic devices. Engineers need to manage heat effectively to prevent overheating and ensure the efficient operation of these systems. For example, in electronic devices like computers and smartphones, heat sinks are used to dissipate heat generated by the components, preventing them from overheating and failing. These are just a few examples of how the concepts of heat transfer and specific heat capacity are applied in real life. From everyday appliances to large-scale industrial processes, understanding these principles is vital for developing efficient and effective technologies.
Practice Questions
To further solidify your understanding of heat transfer and specific heat capacity, working through practice questions is highly beneficial. These questions will help you apply the concepts we've discussed and identify any areas where you may need further review. Let's explore a few practice questions similar to the one we solved earlier. Question 1: How much heat is required to raise the temperature of 250 g of water from 20.0°C to 85.0°C? (Specific heat capacity of water = 1 cal/g°C). This question tests your ability to calculate the heat required to raise the temperature of a given mass of water. You'll need to use the formula Q = m × c × ΔT, where m is the mass, c is the specific heat capacity, and ΔT is the change in temperature. Remember to calculate ΔT correctly by subtracting the initial temperature from the final temperature. Question 2: A 100 g piece of copper at 90.0°C is placed in 50 g of water at 22.0°C. The final temperature of the water and copper is 25.0°C. What is the heat lost by the copper? (Specific heat capacity of copper = 0.093 cal/g°C). This question involves heat transfer between two substances. You'll need to calculate the heat lost by the copper as it cools down. Again, use the formula Q = m × c × ΔT, but be sure to use the correct specific heat capacity for copper. Question 3: If 20,000 calories of heat are added to 1000 g of water at 25.0°C, what will be the final temperature of the water? (Specific heat capacity of water = 1 cal/g°C). This question requires you to rearrange the formula Q = m × c × ΔT to solve for the final temperature. You'll need to calculate ΔT first and then add it to the initial temperature to find the final temperature. Question 4: Calculate the heat released when 800 g of water cools from 75.0°C to 15.0°C. (Specific heat capacity of water = 1 cal/g°C). This question is similar to the original problem we solved. It reinforces the concept of heat release during cooling and the importance of the negative sign in the result. By attempting these practice questions, you'll gain confidence in applying the principles of heat transfer and specific heat capacity. Be sure to show your work and double-check your calculations to avoid common mistakes. Practice is key to mastering these concepts.
Conclusion
In conclusion, determining the amount of heat released when 500 g of water cools from 95.0°C to 25.0°C involves a straightforward application of the specific heat capacity formula. The correct answer, 3.50 x 10^4 cal, is derived by understanding and applying the formula Q = m × c × ΔT. This formula highlights the relationship between heat transfer, mass, specific heat capacity, and temperature change. We calculated the temperature change (ΔT) as -70.0°C and used the specific heat capacity of water (1 cal/g°C) to find the heat released. This problem underscores the fundamental principles of thermochemistry and the importance of accurate calculations in scientific problem-solving. We also discussed common mistakes to avoid, such as incorrectly calculating temperature change, using wrong units, or misapplying the specific heat capacity formula. Being aware of these pitfalls can significantly improve your accuracy in solving similar problems. Furthermore, we explored the real-life applications of heat transfer and specific heat capacity, illustrating how these concepts are crucial in various fields, including heating and cooling systems, the food industry, climate patterns, and engineering. Understanding heat transfer is essential for designing efficient systems, preserving food, and comprehending natural phenomena. Finally, we included several practice questions to help reinforce your understanding and provide opportunities for further learning. These questions challenge you to apply the concepts we've discussed and develop your problem-solving skills. By working through these examples, you can solidify your grasp of heat transfer and specific heat capacity. Mastering these concepts is not only valuable for academic success but also for understanding the world around us. The principles of thermodynamics and heat transfer are fundamental to many aspects of our lives, from the devices we use daily to the climate we experience. Therefore, a solid understanding of these concepts is an invaluable asset.
