Conservation of Energy in Thermal Equilibrium: A Practical Example

Thermal equilibrium is a fascinating aspect of thermodynamics that illustrates the principle of conservation of energy. This article examines a specific problem involving the transfer of heat between a copper block and water. After solving the problem, we will see how the principle of conservation of energy is applied in real-world scenarios.

Problem Statement

Consider a copper block of mass 40 grams at 200°C, which is transferred into a copper container of 60 grams, containing 10 grams of water at 25°C. Assuming no heat is lost to the surroundings, what is the final temperature of the system?

Given Data

Mass of hot copper, mCu1 40 g Initial temperature of hot copper, TCu1 200°C Mass of cold copper, mCu2 60 g Initial temperature of cold copper, TCu2 25°C Mass of water, mwater 10 g Initial temperature of water, Twater 25°C Specific heat capacity of copper, cCu 0.4 J/g°C Specific heat capacity of water, cwater 4.2 J/g°C

Solution

According to the principle of conservation of energy, the heat lost by the hot copper will be equal to the heat gained by the cold copper and the water. Let Tf be the final temperature of the system.

Heat Lost by the Hot Copper

The equation for the heat lost by the hot copper is:

Qlost mCu1 × cCu × (TCu1 - Tf)

Substituting the given values:

Qlost 40 g × 0.4 J/g°C × (200°C - Tf)

Heat Gained by the Cold Copper and Water

The equation for the heat gained by the cold copper and the water is:

Qgained (mCu2 × cCu × (Tf - TCu2) mwater × cwater × (Tf - Twater))

Substituting the given values:

Qgained (60 g × 0.4 J/g°C × (Tf - 25°C) 10 g × 4.2 J/g°C × (Tf - 25°C))

Setting Heat Lost Equal to Heat Gained

Setting Qlost equal to Qgained:

40 g × 0.4 J/g°C × (200°C - Tf) (60 g × 0.4 J/g°C × (Tf - 25°C) 10 g × 4.2 J/g°C × (Tf - 25°C))

Expanding and simplifying:

16 g × 200°C - 16Tf 24Tf - 600 42Tf - 1005

Combining like terms:

3200 - 16Tf 66Tf - 1605

Reorganizing the equation:

3200 1605 66Tf 16Tf

4805 82Tf

Tf 4805 / 82 ≈ 59.15°C

The final temperature Tf of the system is approximately 59.15°C.

Practical Application

This scenario demonstrates the principle of conservation of energy in thermal equilibrium, where the heat lost by the hotter object is exactly equal to the heat gained by the colder objects involved. This is a fundamental concept in thermodynamics and has numerous practical applications in fields such as engineering, HVAC, and chemical processes.

Conclusion

Understanding the conservation of energy in thermal equilibrium is crucial in many real-world applications. By solving practical problems like the one presented, we can better understand how heat transfer works and apply this knowledge to design efficient systems and processes.