Calculating Final Temperature of Two Solids in Thermal Contact

In summary, the problem involves two solids, A and B, with different properties, being placed in thermal contact and isolated. The goal is to find the final temperature of both solids after solid A completely melts. Using the equations Q=mL+mCdeltaT and Q=mCBTfinal-mCBTB, and the fact that the two solids are thermally isolated, we can determine that the final temperature, Tf, is equal to (MCATA+mCBTB-ML)/(MCA+mCB). The "m" terms cancel out in this equation, allowing us to solve for Tf. It is important to have a solid understanding of the concepts involved in this problem before attempting to solve it.
  • #1
Any Help
79
2

Homework Statement


55. Solid A, with mass M, is at its melting point TA. It is placed in thermal contact with solid B, with heat capacity CB and initially at temperature TB (TB > TA). The combination is thermally isolated. A has latent heat of fusion L and when it has melted has heat capacity CA. If A completely melts the final temperature of both A and B is:
A. (CATA + CBTB − ML)/(CA + CB)
B. (CATA − CBTB + ML)/(CA + CB)
C. (CATA − CBTB − ML)/(CA + CB)
D. (CATA + CBTB + ML)/(CA − CB)
E. (CATA + CBTB + ML)/(CA − CB)
correct answer is A ?

Homework Equations


Q=mL+mCdeltaT

The Attempt at a Solution


in A Q1=ML+MCATfinal-MCATA
in B Q=mCBTfinal-mCBTB
-Q1=Q
-ML-MCATfinal+MCATA=mCBTfinal-mCBTB
-ML+MCATA+mCNTB=MCATf+mCBTf
Tf=(MCATA+mCBTB-ML)/(MCA+mCB)
how they get rid of the m in the above equation?
 
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  • #2
Q=mL=mCdeltaT
... this is not strictly correct because mL is not usually going to be equal to mCdeltaT.

how they get rid of the m in the above equation?
... you need to explain the reasoning (physics) behind your maths, but basically, in one of the steps, the "m" terms cancel out.
Try dividing the problem into two stages. Take the algebra carefully, step by step, and explain each step you do.
 
  • #3
Simon Bridge said:
Q=mL=mCdeltaT
sorry it is a keyboard mistake I mean Q=mL+mCdeltaT
 
  • #4
Any Help said:
sorry it is a keyboard mistake I mean Q=mL+mCdeltaT
Um OK. So what is your reasoning?

Note: you should probably get the concepts down from the other question before tackling this one.
 

Related to Calculating Final Temperature of Two Solids in Thermal Contact

What is the first law of thermodynamics?

The first law of thermodynamics, also known as the law of conservation of energy, states that energy cannot be created or destroyed, only transferred or converted from one form to another.

What is the significance of the first law of thermodynamics?

The first law of thermodynamics is significant because it helps us understand the relationship between different forms of energy and how energy is conserved in a closed system. It also lays the foundation for the study of thermodynamics and its applications in various fields of science and engineering.

How does the first law of thermodynamics apply to everyday life?

The first law of thermodynamics applies to everyday life in many ways. For example, when we turn on a light bulb, the electrical energy is converted into light and heat energy. When we eat food, our bodies convert the chemical energy in the food into kinetic energy for movement and other bodily functions. Essentially, all energy transformations in our daily lives follow the first law of thermodynamics.

Can the first law of thermodynamics be violated?

No, the first law of thermodynamics is a fundamental law of nature and cannot be violated. The total amount of energy in a closed system remains constant, and energy cannot be created or destroyed. However, it can be transformed from one form to another.

How is the first law of thermodynamics related to the second law?

The first law of thermodynamics deals with the conservation of energy, while the second law of thermodynamics deals with the direction of energy flow and the increase of entropy in a closed system. Both laws are fundamental in understanding the behavior of energy in different systems and are interconnected in their applications.

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