Calculate Tin Sample Mass Using Specific Heat Capacity A Step-by-Step Guide

In the realm of thermodynamics, understanding the relationship between heat transfer, temperature change, and mass is crucial. The specific heat capacity, a fundamental property of materials, plays a pivotal role in these calculations. This article delves into a practical problem involving heat transfer in a tin sample, demonstrating how to calculate the mass of the sample using the principles of specific heat capacity. We will explore the underlying concepts, walk through the step-by-step solution, and highlight the significance of this calculation in various scientific and engineering applications.

Specific heat capacity is a material property that quantifies the amount of heat energy required to raise the temperature of one gram of a substance by one degree Celsius (or one Kelvin). It is represented by the symbol Cp and is typically measured in joules per gram per degree Celsius (J/g·°C). Different materials possess different specific heat capacities. For instance, water has a high specific heat capacity (4.186 J/g·°C), meaning it requires a significant amount of energy to change its temperature, while metals generally have lower specific heat capacities. Understanding specific heat capacity is essential in various fields, including materials science, engineering, and chemistry, as it helps predict how a material will respond to changes in temperature and energy transfer. The formula that relates heat transfer (q), mass (m), specific heat capacity (Cp), and temperature change (ΔT) is:

q = m * Cp * ΔT

Where:

• q is the heat transferred (in joules)
• m is the mass of the substance (in grams)
• Cp is the specific heat capacity of the substance (in J/g·°C)
• ΔT is the change in temperature (in °C), calculated as the final temperature minus the initial temperature.

The specific heat capacity is an intensive property, meaning it does not depend on the amount of substance. This value is unique to each substance and is a crucial factor in determining how much energy is needed to change its temperature. For example, water has a high specific heat capacity compared to most metals, which explains why it takes more energy to heat up water. This property is essential in many applications, such as in cooling systems, where water is used to absorb heat due to its high heat capacity. Furthermore, the specific heat capacity of a substance can vary slightly with temperature and phase. However, for many practical applications, assuming a constant specific heat capacity over a moderate temperature range is a good approximation. In the context of thermodynamics, the specific heat capacity is a key parameter in calorimetry, where heat transfer is measured and quantified. This measurement allows scientists and engineers to design systems and processes more efficiently, ensuring optimal performance and safety. Understanding the specific heat capacity also helps in material selection for various applications, as it influences how a material stores and releases heat, affecting its thermal behavior.

In this problem, we are presented with a sample of tin, a metal known for its relatively low specific heat capacity. The tin sample is placed in a freezer, causing its temperature to decrease. We are given the specific heat capacity of tin (Cp = 0.227 J/g·°C), the initial temperature (15.0 °C), the final temperature (-10.0 °C), and the amount of energy released (543 J). The objective is to determine the mass of the tin sample. This problem is a classic example of applying the principles of thermodynamics to calculate material properties based on heat transfer measurements. The scenario involves heat flowing out of the tin sample as it cools in the freezer, which is an exothermic process. The negative sign of the heat released indicates that the energy is leaving the system (the tin sample). Understanding the direction of heat flow is crucial in correctly interpreting the problem and applying the appropriate sign conventions in the calculations. The problem also highlights the practical applications of thermodynamics in everyday scenarios, such as refrigeration and cooling processes. By solving this problem, we not only reinforce our understanding of specific heat capacity but also gain insight into how thermal properties affect the behavior of materials in different environments. The problem's straightforward nature makes it an excellent educational tool for students learning about thermodynamics and heat transfer, providing a clear and concise example of how these concepts can be applied to solve real-world problems.

To solve this problem, we will use the formula that relates heat transfer, mass, specific heat capacity, and temperature change:

q = m * Cp * ΔT

Where:

• q = -543 J (heat released, hence the negative sign)
• Cp = 0.227 J/g·°C (specific heat capacity of tin)
• ΔT = Tf - Ti = -10.0 °C - 15.0 °C = -25.0 °C (change in temperature)
• m = mass of the tin sample (what we need to find)

Step 1: Rearrange the formula to solve for mass (m).

We need to isolate m on one side of the equation. Dividing both sides of the equation by Cp and ΔT, we get:

m = q / (Cp * ΔT)

Step 2: Substitute the given values into the rearranged formula.

Now, we plug in the values we have:

m = -543 J / (0.227 J/g·°C * -25.0 °C)

Step 3: Perform the calculation.

First, multiply the values in the denominator:

0. 227 J/g·°C * -25.0 °C = -5.675 J/g

Then, divide the heat released by this value:

m = -543 J / -5.675 J/g = 95.7 g

Step 4: Round the answer to three significant figures.

The question asks us to round the answer to three significant figures. Therefore,

m ≈ 95.7 g

The step-by-step solution provided is a systematic approach to solving the problem, ensuring that each parameter is correctly substituted and calculated. The rearrangement of the formula is a critical step, showcasing the algebraic manipulation required in thermodynamic calculations. The inclusion of the negative sign for the heat released is essential to reflect the exothermic nature of the process, where energy is leaving the system. The calculation itself involves basic arithmetic operations, but the accuracy of the final result depends on the precision of the input values and the correct application of the formula. Rounding the answer to the appropriate number of significant figures is a crucial step in scientific calculations, as it reflects the precision of the measurements and calculations involved. The final answer, 95.7 g, represents the mass of the tin sample that released 543 J of energy while cooling from 15.0 °C to -10.0 °C. This result is consistent with the given specific heat capacity of tin, providing a practical application of the principles of thermodynamics in determining material properties.

To fully grasp the solution, it's essential to understand the underlying concepts and how they are applied in the problem. The main concepts are heat transfer, specific heat capacity, and temperature change.

Heat Transfer:

Heat transfer is the process by which thermal energy moves from one system to another. In this case, the tin sample releases heat into the freezer, which acts as a heat sink. The amount of heat transferred (q) is a crucial parameter in our calculation. Because the tin is releasing energy, the value of q is negative (-543 J). This negative sign is critical because it indicates the direction of heat flow. If the tin were absorbing heat, q would be positive. Understanding the sign convention for heat transfer is vital in thermodynamics, as it directly affects the interpretation of the results. Heat transfer can occur through various mechanisms, including conduction, convection, and radiation. In this scenario, heat is likely transferred through a combination of these mechanisms, but the net effect is that the tin loses thermal energy to the colder environment of the freezer. The amount of heat transferred depends on several factors, including the temperature difference between the tin and the freezer, the thermal conductivity of the materials involved, and the surface area available for heat transfer. In the context of this problem, we are given the total amount of heat transferred, which simplifies the calculation by allowing us to focus on the relationship between heat transfer, mass, specific heat capacity, and temperature change.

Specific Heat Capacity:

Specific heat capacity (Cp) is a measure of how much energy is required to change the temperature of a substance. Tin has a specific heat capacity of 0.227 J/g·°C, which means it takes 0.227 joules of energy to raise the temperature of 1 gram of tin by 1 degree Celsius. Different materials have different specific heat capacities. For example, water has a much higher specific heat capacity (4.186 J/g·°C) than tin, meaning it takes significantly more energy to heat water compared to tin. This difference in specific heat capacity explains why water is often used as a coolant in various applications. The specific heat capacity is an intrinsic property of a substance, meaning it is a characteristic of the material itself and does not depend on the amount of substance present. However, the total amount of heat required to change the temperature of a substance does depend on its mass, which is why mass is a crucial parameter in our calculation. Understanding specific heat capacity is essential in various fields, including materials science, engineering, and chemistry, as it helps predict how a material will respond to changes in temperature and energy transfer. In the context of this problem, the specific heat capacity of tin is a key piece of information that allows us to relate the heat released to the temperature change and the mass of the sample.

Temperature Change:

Temperature change (ΔT) is the difference between the final temperature and the initial temperature. In this problem, the tin cools from 15.0 °C to -10.0 °C, so the temperature change is -25.0 °C. The negative sign indicates that the temperature decreased. The temperature change is a critical factor in determining the amount of heat transferred. A larger temperature change implies that more heat has been transferred, assuming the mass and specific heat capacity remain constant. In thermodynamic calculations, it is essential to use consistent units for temperature. While specific heat capacity is often given in J/g·°C, the temperature change can also be expressed in Kelvin (K), since a change of 1 degree Celsius is equivalent to a change of 1 Kelvin. However, in this problem, since we are dealing with a temperature difference, using Celsius is perfectly acceptable. The temperature change also provides information about the direction of heat flow. A negative temperature change, as in this case, indicates that heat is being released from the system, while a positive temperature change would indicate that heat is being absorbed. Understanding the relationship between temperature change and heat transfer is fundamental to solving thermodynamic problems and interpreting the results accurately.

The concept of specific heat capacity is not just a theoretical construct; it has numerous practical applications in various fields, including:

1. Engineering: In mechanical engineering, specific heat capacity is crucial in designing cooling systems for engines and machinery. Materials with high specific heat capacities, like water, are often used as coolants because they can absorb a large amount of heat without a significant temperature increase. This property helps prevent overheating and ensures the efficient operation of mechanical systems. In civil engineering, specific heat capacity is considered in the design of buildings and infrastructure. For example, materials with high specific heat capacities can help regulate indoor temperatures, reducing the need for excessive heating or cooling. In electrical engineering, specific heat capacity is important in managing the thermal behavior of electronic components, preventing them from overheating and ensuring their reliable performance.

2. Materials Science: Specific heat capacity is a key parameter in materials science for characterizing the thermal behavior of different substances. It helps in selecting appropriate materials for various applications, such as heat shields, insulators, and heat sinks. For instance, materials with low specific heat capacities heat up quickly and are suitable for applications where rapid temperature changes are required. Conversely, materials with high specific heat capacities are used in applications where thermal stability is essential. Understanding the specific heat capacity of materials is also crucial in designing thermal processing techniques, such as heat treatment and annealing, which are used to modify the properties of materials.

3. Chemistry: In chemistry, specific heat capacity is used in calorimetry, a technique for measuring heat transfer in chemical reactions. By measuring the temperature change in a calorimeter, chemists can determine the amount of heat released or absorbed in a reaction, which is essential for understanding the reaction's thermodynamics. Specific heat capacity is also important in understanding the thermal behavior of solutions and mixtures, as it affects their heating and cooling characteristics. Furthermore, specific heat capacity plays a role in chemical processes such as distillation and evaporation, where heat transfer is a critical factor.

4. Cooking: Specific heat capacity plays a significant role in cooking. For example, water's high specific heat capacity allows it to heat food evenly and efficiently. Different cooking methods, such as boiling, steaming, and frying, rely on the specific heat capacities of various substances to transfer heat to the food. Understanding specific heat capacity can also help in selecting appropriate cookware. For example, pots and pans made of materials with high thermal conductivity and appropriate specific heat capacities ensure even heat distribution and prevent hot spots.

5. Climate Science: Specific heat capacity is a crucial factor in climate science, particularly in understanding the Earth's climate system. Water's high specific heat capacity plays a significant role in regulating global temperatures. Oceans, which cover a large portion of the Earth's surface, absorb and store vast amounts of heat, moderating temperature fluctuations and influencing weather patterns. The specific heat capacity of land surfaces also affects regional climates, with areas having higher heat capacities experiencing smaller temperature variations. Understanding these thermal properties is essential for modeling climate change and predicting its impacts.

In conclusion, determining the mass of the tin sample using specific heat capacity involves understanding the relationship between heat transfer, mass, specific heat capacity, and temperature change. By applying the formula q = m * Cp * ΔT and rearranging it to solve for mass, we successfully calculated the mass of the tin sample to be approximately 95.7 grams. This problem illustrates the practical application of thermodynamics in determining material properties based on heat transfer measurements. The understanding of specific heat capacity is paramount in many scientific and engineering disciplines, and its applications span across various real-world scenarios. From designing efficient cooling systems to understanding climate patterns, the principles of specific heat capacity provide a fundamental framework for analyzing and predicting thermal behavior. The ability to solve such problems not only reinforces theoretical knowledge but also provides valuable insights into the practical implications of thermodynamic principles. The step-by-step solution presented in this article serves as a guide for approaching similar problems, emphasizing the importance of careful consideration of units, sign conventions, and the accurate application of formulas. By mastering these concepts, students and professionals can effectively tackle a wide range of problems involving heat transfer and material properties, contributing to advancements in various fields.