A Sample Of Tin \[$\left( C_p = 0.227 \, \text{J/g} \cdot {}^{\circ} \text{C} \right)\$\] Is Placed In A Freezer. Its Temperature Decreases From \[$15.0^{\circ} \text{C}\$\] To \[$-10.0^{\circ} \text{C}\$\] As It Releases 513 J
Introduction
In this article, we will explore the cooling process of a sample of tin, focusing on the heat released as its temperature decreases. The specific heat capacity of tin is given as 0.227 J/g°C, and we will use this value to calculate the heat released during the cooling process.
The Cooling Process
When a sample of tin is placed in a freezer, its temperature decreases as it releases heat to the surroundings. The heat released is directly proportional to the change in temperature of the sample. We can use the formula for heat transfer to calculate the heat released:
Q = mcΔT
where Q is the heat released, m is the mass of the sample, c is the specific heat capacity, and ΔT is the change in temperature.
Calculating the Heat Released
Given that the temperature of the tin sample decreases from 15.0°C to -10.0°C, we can calculate the change in temperature as follows:
ΔT = T_initial - T_final = 15.0°C - (-10.0°C) = 25.0°C
We are also given that the heat released is 513 J. We can use the formula for heat transfer to calculate the mass of the sample:
m = Q / (cΔT)
Substituting the given values, we get:
m = 513 J / (0.227 J/g°C × 25.0°C) = 90.9 g
Understanding the Cooling Process
The cooling process of the tin sample can be understood by analyzing the heat transfer that occurs between the sample and the surroundings. As the sample cools, it releases heat to the surroundings, causing its temperature to decrease. The rate of heat transfer is directly proportional to the temperature difference between the sample and the surroundings.
Factors Affecting the Cooling Process
Several factors can affect the cooling process of the tin sample, including:
- Mass of the sample: A larger mass of the sample will result in a greater heat capacity, causing the sample to cool more slowly.
- Specific heat capacity: A higher specific heat capacity will result in a greater heat capacity, causing the sample to cool more slowly.
- Temperature difference: A larger temperature difference between the sample and the surroundings will result in a greater rate of heat transfer.
- Heat transfer coefficient: A higher heat transfer coefficient will result in a greater rate of heat transfer.
Conclusion
In conclusion, the cooling process of a sample of tin can be understood by analyzing the heat transfer that occurs between the sample and the surroundings. The heat released is directly proportional to the change in temperature of the sample, and the mass of the sample, specific heat capacity, temperature difference, and heat transfer coefficient all play a role in determining the rate of heat transfer.
References
- Specific heat capacity of tin: 0.227 J/g°C
- Heat released: 513 J
- Temperature change: 25.0°C
Further Reading
For further information on the cooling process of tin, we recommend the following resources:
- Thermodynamics: A comprehensive textbook on thermodynamics, covering the principles of heat transfer and thermodynamic systems.
- Heat transfer: A detailed article on heat transfer, covering the principles of heat transfer and its applications.
- Specific heat capacity: A detailed article on specific heat capacity, covering the definition, units, and applications of specific heat capacity.
A Sample of Tin: Q&A =========================
Frequently Asked Questions
Q: What is the specific heat capacity of tin? A: The specific heat capacity of tin is 0.227 J/g°C.
Q: How much heat is released by a sample of tin as it cools from 15.0°C to -10.0°C? A: The heat released by a sample of tin as it cools from 15.0°C to -10.0°C is 513 J.
Q: What is the change in temperature of the sample? A: The change in temperature of the sample is 25.0°C.
Q: How can we calculate the mass of the sample? A: We can calculate the mass of the sample using the formula:
m = Q / (cΔT)
where m is the mass of the sample, Q is the heat released, c is the specific heat capacity, and ΔT is the change in temperature.
Q: What factors affect the cooling process of the sample? A: Several factors can affect the cooling process of the sample, including:
- Mass of the sample: A larger mass of the sample will result in a greater heat capacity, causing the sample to cool more slowly.
- Specific heat capacity: A higher specific heat capacity will result in a greater heat capacity, causing the sample to cool more slowly.
- Temperature difference: A larger temperature difference between the sample and the surroundings will result in a greater rate of heat transfer.
- Heat transfer coefficient: A higher heat transfer coefficient will result in a greater rate of heat transfer.
Q: What is the significance of the specific heat capacity of tin? A: The specific heat capacity of tin is an important property that determines how much heat is required to change the temperature of the sample. A higher specific heat capacity means that more heat is required to change the temperature of the sample.
Q: How can we apply the principles of heat transfer to real-world problems? A: The principles of heat transfer can be applied to a wide range of real-world problems, including:
- Designing cooling systems: Understanding the principles of heat transfer can help us design more efficient cooling systems for buildings, vehicles, and other applications.
- Improving energy efficiency: By optimizing heat transfer, we can reduce energy consumption and improve energy efficiency in a wide range of applications.
- Developing new materials: Understanding the principles of heat transfer can help us develop new materials with improved thermal properties.
Q: What are some common applications of heat transfer? A: Heat transfer has a wide range of applications, including:
- Cooling systems: Heat transfer is used in cooling systems for buildings, vehicles, and other applications.
- Refrigeration: Heat transfer is used in refrigeration systems to cool food and other products.
- Power generation: Heat transfer is used in power generation systems to convert heat into electricity.
- Materials processing: Heat transfer is used in materials processing to heat and cool materials for a wide range of applications.
Q: What are some common misconceptions about heat transfer? A: Some common misconceptions about heat transfer include:
- Heat transfer is only important at high temperatures: Heat transfer is important at all temperatures, from very low temperatures to very high temperatures.
- Heat transfer is only important in certain applications: Heat transfer is important in a wide range of applications, from cooling systems to power generation.
- Heat transfer is only related to temperature: Heat transfer is related to temperature, but it is also related to other factors, including heat capacity, thermal conductivity, and heat transfer coefficient.