Zoe Has 25 Grams Of Water { \left(c=4.186 \frac{J}{g^{\circ} C }\right)$}$ At ${ 10^{\circ} C\$} , Which She Mixes With 12 Grams Of Water At ${ 30^{\circ} C\$} . Assume That No Heat Is Lost To The Surroundings.What Is The
Introduction
Heat transfer is a fundamental concept in physics that plays a crucial role in various natural phenomena and engineering applications. In this article, we will explore a real-world scenario involving heat transfer and temperature equilibrium. Zoe has 25 grams of water at 10°C, which she mixes with 12 grams of water at 30°C. We will assume that no heat is lost to the surroundings and determine the final temperature of the mixture.
The Problem
Zoe has 25 grams of water at 10°C, which she mixes with 12 grams of water at 30°C. The specific heat capacity of water is given as 4.186 J/g°C. We need to find the final temperature of the mixture after the two water samples are mixed together.
Assumptions
- No heat is lost to the surroundings.
- The specific heat capacity of water is constant at 4.186 J/g°C.
- The mixture is homogeneous, meaning that the temperature is uniform throughout.
Heat Transfer Equations
The heat transfer equation for a system is given by:
Q = mcΔT
where Q is the heat transferred, m is the mass of the system, c is the specific heat capacity, and ΔT is the change in temperature.
Since no heat is lost to the surroundings, the heat transferred from the hotter water to the cooler water is equal to the heat transferred from the cooler water to the hotter water. We can set up two heat transfer equations:
Q1 = m1cΔT1 Q2 = m2cΔT2
where Q1 and Q2 are the heat transferred from the hotter water to the cooler water and from the cooler water to the hotter water, respectively.
Solving for the Final Temperature
We can set up an equation using the heat transfer equations:
m1cΔT1 = m2cΔT2
Substituting the given values, we get:
25g x 4.186 J/g°C x (Tf - 10°C) = 12g x 4.186 J/g°C x (30°C - Tf)
where Tf is the final temperature of the mixture.
Simplifying the equation, we get:
104.65 Tf - 258.5 = 50.112 - 50.112 Tf
Combine like terms:
154.762 Tf = 308
Divide by 154.762:
Tf = 2°C
Conclusion
In this article, we explored a real-world scenario involving heat transfer and temperature equilibrium. We assumed that no heat is lost to the surroundings and determined the final temperature of the mixture. The final temperature of the mixture is 2°C.
Key Takeaways
- Heat transfer is a fundamental concept in physics that plays a crucial role in various natural phenomena and engineering applications.
- The heat transfer equation is given by Q = mcΔT.
- The specific heat capacity of water is constant at 4.186 J/g°C.
- The mixture is homogeneous, meaning that the temperature is uniform throughout.
- The final temperature of the mixture is 2°C.
Real-World Applications
Heat transfer is a crucial concept in various real-world applications, including:
- Refrigeration: Heat transfer is used to transfer heat from the cold interior of a refrigerator to the hot exterior.
- Air Conditioning: Heat transfer is used to transfer heat from the hot interior of a building to the cool exterior.
- Power Generation: Heat transfer is used to transfer heat from the hot interior of a power plant to the cool exterior.
- Food Processing: Heat transfer is used to transfer heat from the hot interior of a food processor to the cool exterior.
Future Research Directions
Heat transfer is a complex phenomenon that is still not fully understood. Future research directions include:
- Investigating the effects of turbulence on heat transfer: Turbulence can significantly affect heat transfer rates.
- Developing new materials with high thermal conductivity: New materials with high thermal conductivity can improve heat transfer rates.
- Investigating the effects of surface roughness on heat transfer: Surface roughness can significantly affect heat transfer rates.
References
- Cengel, Y. A. (2018). Heat Transfer: A Practical Approach. McGraw-Hill Education.
- Incropera, F. P., & Dewitt, D. P. (2017). Fundamentals of Heat and Mass Transfer. John Wiley & Sons.
- Kittel, C. (2005). Introduction to Solid State Physics. John Wiley & Sons.
Heat Transfer and Temperature Equilibrium: A Real-World Scenario - Q&A ====================================================================
Introduction
In our previous article, we explored a real-world scenario involving heat transfer and temperature equilibrium. We assumed that no heat is lost to the surroundings and determined the final temperature of the mixture. In this article, we will answer some frequently asked questions related to heat transfer and temperature equilibrium.
Q: What is heat transfer?
A: Heat transfer is the transfer of thermal energy from one body or system to another due to a temperature difference. It is a fundamental concept in physics that plays a crucial role in various natural phenomena and engineering applications.
Q: What are the three modes of heat transfer?
A: The three modes of heat transfer are:
- Conduction: Heat transfer through direct contact between particles or molecules.
- Convection: Heat transfer through the movement of fluids.
- Radiation: Heat transfer through electromagnetic waves.
Q: What is the specific heat capacity of water?
A: The specific heat capacity of water is 4.186 J/g°C. This means that it takes 4.186 joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.
Q: How do you calculate the heat transferred in a system?
A: The heat transferred in a system can be calculated using the equation:
Q = mcΔT
where Q is the heat transferred, m is the mass of the system, c is the specific heat capacity, and ΔT is the change in temperature.
Q: What is the final temperature of the mixture in the scenario we discussed earlier?
A: The final temperature of the mixture is 2°C.
Q: Why is it important to consider the specific heat capacity of a substance when calculating heat transfer?
A: The specific heat capacity of a substance is important because it determines how much energy is required to raise the temperature of a given mass of the substance by a certain amount. If the specific heat capacity is high, more energy is required to raise the temperature, and vice versa.
Q: Can heat transfer occur without a temperature difference?
A: No, heat transfer cannot occur without a temperature difference. Heat transfer is a spontaneous process that occurs from a region of higher temperature to a region of lower temperature.
Q: What are some real-world applications of heat transfer?
A: Some real-world applications of heat transfer include:
- Refrigeration: Heat transfer is used to transfer heat from the cold interior of a refrigerator to the hot exterior.
- Air Conditioning: Heat transfer is used to transfer heat from the hot interior of a building to the cool exterior.
- Power Generation: Heat transfer is used to transfer heat from the hot interior of a power plant to the cool exterior.
- Food Processing: Heat transfer is used to transfer heat from the hot interior of a food processor to the cool exterior.
Q: What are some future research directions in heat transfer?
A: Some future research directions in heat transfer include:
- Investigating the effects of turbulence on heat transfer: Turbulence can significantly affect heat transfer rates.
- Developing new materials with high thermal conductivity: New materials with high thermal conductivity can improve heat transfer rates.
- Investigating the effects of surface roughness on heat transfer: Surface roughness can significantly affect heat transfer rates.
Conclusion
In this article, we answered some frequently asked questions related to heat transfer and temperature equilibrium. We hope that this article has provided a better understanding of the concept of heat transfer and its importance in various real-world applications.