A 51.6 G Sample Of Brass Is Put Into A Calorimeter That Contains 200.0 G Of Water. The Brass Sample Starts Off At 92.2 ∘ C 92.2{ }^{\circ} C 92.2 ∘ C , And The Temperature Of The Water Starts Off At 21.0 ∘ C 21.0^{\circ} C 21. 0 ∘ C . When The Temperature Of The Water

Introduction

In this article, we will delve into the world of thermodynamics and explore the concept of heat transfer. Specifically, we will examine the scenario where a 51.6 g sample of brass is placed in a calorimeter containing 200.0 g of water. The brass sample starts at a temperature of $92.2{ }^{\circ} C$, while the water is at $21.0^{\circ} C$. As the brass sample and water interact, heat is transferred from the brass to the water, causing the temperature of the water to rise. In this discussion, we will analyze the heat transfer process and calculate the final temperature of the water.

The Calorimeter and Its Components

A calorimeter is a device used to measure the heat of a chemical reaction or the heat transferred between two substances. In this scenario, the calorimeter contains 200.0 g of water, which serves as the heat sink. The brass sample, with a mass of 51.6 g, is placed in the calorimeter and is the source of heat.

Heat Transfer and the First Law of Thermodynamics

Heat transfer occurs when there is a temperature difference between two substances. In this case, the brass sample is at a higher temperature than the water, causing heat to flow from the brass to the water. The first law of thermodynamics states that energy cannot be created or destroyed, only converted from one form to another. In the context of heat transfer, this means that the energy lost by the brass sample is equal to the energy gained by the water.

Calculating the Heat Transfer

To calculate the heat transfer, we need to use the formula:

Q = mcΔT

where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

For the brass sample, the specific heat capacity is 0.385 J/g°C, and the initial temperature is $92.2{ }^{\circ} C$. The final temperature of the brass sample is unknown, but we can assume it will be equal to the final temperature of the water.

For the water, the specific heat capacity is 4.184 J/g°C, and the initial temperature is $21.0^{\circ} C$. The final temperature of the water is also unknown.

Setting Up the Equations

We can set up two equations using the formula Q = mcΔT:

For the brass sample:

Q = (51.6 g)(0.385 J/g°C)(Tf - 92.2°C)

For the water:

Q = (200.0 g)(4.184 J/g°C)(Tf - 21.0°C)

where Tf is the final temperature of the brass sample and the water.

Solving the Equations

Since the heat transferred from the brass sample is equal to the heat gained by the water, we can set the two equations equal to each other:

(51.6 g)(0.385 J/g°C)(Tf - 92.2°C) = (200.0 g)(4.184 J/g°C)(Tf - 21.0°C)

Simplifying the equation, we get:

19.9 Tf - 1803.3 = 836.8 Tf - 8680

Combine like terms:

856.7 Tf = 6887.3

Divide by 856.7:

Tf = 8.03°C

Conclusion

In this discussion, we analyzed the heat transfer process between a 51.6 g sample of brass and 200.0 g of water in a calorimeter. We calculated the final temperature of the water using the first law of thermodynamics and the formula Q = mcΔT. The final temperature of the water was found to be 8.03°C.

References

• [1] Hall, J. D. (2019). Thermodynamics: An Introduction to the Physical Theories of Equilibrium Thermostatics and Irreversible Thermodynamics. Cambridge University Press.
• [2] Atkins, P. W. (2018). Physical Chemistry. Oxford University Press.

Additional Resources

• [1] Khan Academy. (n.d.). Thermodynamics. Retrieved from https://www.khanacademy.org/science/chemistry/thermodynamics
• [2] MIT OpenCourseWare. (n.d.). 5.60 Thermodynamics. Retrieved from https://ocw.mit.edu/courses/chemistry-and-biology/5-60-thermodynamics-fall-2008/
  A 51.6 g Sample of Brass in a Calorimeter: Q&A =====================================================

Q: What is the purpose of a calorimeter in this experiment?

A: The purpose of a calorimeter in this experiment is to measure the heat transfer between the brass sample and the water. The calorimeter allows us to contain the water and brass sample in a controlled environment, making it easier to measure the heat transfer.

Q: What is the specific heat capacity of brass and water?

A: The specific heat capacity of brass is 0.385 J/g°C, and the specific heat capacity of water is 4.184 J/g°C.

Q: How do we calculate the heat transfer between the brass sample and the water?

A: We use the formula Q = mcΔT to calculate the heat transfer. Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

Q: What is the initial temperature of the brass sample and the water?

A: The initial temperature of the brass sample is $92.2{ }^{\circ} C$, and the initial temperature of the water is $21.0^{\circ} C$.

Q: What is the final temperature of the brass sample and the water?

A: The final temperature of the brass sample and the water is 8.03°C.

Q: Why is the final temperature of the brass sample and the water the same?

A: The final temperature of the brass sample and the water is the same because the heat transferred from the brass sample is equal to the heat gained by the water. This is a result of the first law of thermodynamics, which states that energy cannot be created or destroyed, only converted from one form to another.

Q: What is the significance of this experiment?

A: This experiment demonstrates the concept of heat transfer and the first law of thermodynamics. It shows how heat can be transferred from one substance to another, and how the temperature of a substance can change as a result of heat transfer.

Q: What are some real-world applications of this concept?

A: There are many real-world applications of this concept, including:

• Heating and cooling systems: Heat transfer is used in heating and cooling systems to transfer heat from one location to another.
• Refrigeration: Heat transfer is used in refrigeration systems to transfer heat from a cold location to a warm location.
• Power generation: Heat transfer is used in power generation systems to transfer heat from a hot location to a cold location.
• Materials science: Heat transfer is used in materials science to study the thermal properties of materials.

Q: What are some common mistakes to avoid when performing this experiment?

A: Some common mistakes to avoid when performing this experiment include:

• Incorrect measurement of the mass and temperature of the substances: Make sure to accurately measure the mass and temperature of the substances.
• Incorrect calculation of the heat transfer: Make sure to accurately calculate the heat transfer using the formula Q = mcΔT.
• Incorrect handling of the calorimeter: Make sure to handle the calorimeter carefully to avoid any damage or contamination.

Q: What are some tips for improving the accuracy of this experiment?

A: Some tips for improving the accuracy of this experiment include:

• Use high-quality equipment: Use high-quality equipment, such as a precise thermometer and a accurate balance.
• Follow proper procedures: Follow proper procedures, such as accurately measuring the mass and temperature of the substances and accurately calculating the heat transfer.
• Repeat the experiment multiple times: Repeat the experiment multiple times to ensure accurate results.