Thermal physics final temperature of mixture

In summary, the problem involves finding the final temperature of 250.0g of copper at 100.0°C and 325.0g of water at 20.0°C, assuming no heat loss to the surroundings. The equation used is mcCΔTc + mhCΔTh = 0, and the solution involves setting up the equation properly by taking into account the direction of heat flow and using the correct signs for ΔT. Conservation of energy is used in all problems involving heat flow between two materials.
  • #1
AbsoluteZer0
125
1

Homework Statement




250.0g of copper at 100.0°C are placed in a cup containing 325.0g of water at 20.0°C. Assume no heat loss to the surroundings. What is the final temperature of the copper and water?

Homework Equations



Conservation of Energy
mcCΔTc = mhCΔTh

The Attempt at a Solution



mcCΔTc = mhCΔTh
(0.250)(390)(Tf - Ti) = (0.325)(4200)(Tf - Ti)
(0.250)(390)(Tf - 100) = (0.325)(4200)(Tf - 20)

Have I set up my equation properly?
 
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  • #2
AbsoluteZer0 said:

Homework Statement

250.0g of copper at 100.0°C are placed in a cup containing 325.0g of water at 20.0°C. Assume no heat loss to the surroundings. What is the final temperature of the copper and water?

Homework Equations



Conservation of Energy
mcCΔTc = mhCΔTh

The Attempt at a Solution



mcCΔTc = mhCΔTh
(0.250)(390)(Tf - Ti) = (0.325)(4200)(Tf - Ti)
(0.250)(390)(Tf - 100) = (0.325)(4200)(Tf - 20)

Have I set up my equation properly?
Almost, but not quite. Notice that you have a negative quantity on the left hand side, and a positive quantity on the right hand side -- because Tf must be somewhere between 20 and 100 C, right? So something is definitely wrong here.

Let's back up to your starting equation, which really should say

mcCcΔTc + mhChΔTh = 0

(Conservation of energy means that the total change in energy is zero.)
 
  • #3
mcCcΔTc + mhChΔTh = 0

(Conservation of energy means that the total change in energy is zero.)

Thanks

Hypothetically speaking, if the question asked for the the final temperature of the Copper would the set up that I used be appropriate had the final temperature of water been given?
 
  • #4
AbsoluteZer0 said:
Thanks

Hypothetically speaking, if the question asked for the the final temperature of the Copper would the set up that I used be appropriate had the final temperature of water been given?
They would never ask it that way, since the copper and water have the same final temperature. Or did you mean to say something else?

As for your original equation, let's take a look at it again:
[tex]m_cC_c \Delta T_c = m_h C_h \Delta T_h[/tex]
Think about what it is really claiming -- it is claiming that the same amount of energy enters both the copper and the water (imagine that the ΔT's are both positive). Where is that energy supposed to come from? This claim just doesn't make sense, and the equation not saying that energy is conserved as was your intention in your original post.

We need to have a minus sign on one side of that equation. Then it would be saying the energy that leaves one substance equals the energy that enters the other substance. In other words, some energy can move from one substance to the other, and none of the energy can simply disappear, or appear out of nowhere. That is what conservation of energy means.

Note, in some examples worked out in a textbook or class lectures, people might define the ΔT's differently and write the equation you wrote. That is, for the hotter substance (copper here) they might really mean [itex]\Delta T = T_i-T_f[/itex], which is the opposite of what you called it, and equivalent to putting a minus sign in your equation as required.

Hope that helps clear things up.
 
  • #5
So we use

[itex] m_c C_c \Delta T_c + m_h C_h \Delta T_h = 0 [/itex]

In all questions regarding conservation of energy?
 
  • #6
In the ones involving heat flow between two materials, yes.

(Conservation of energy is also used in questions dealing with kinetic and potential energy. A different equation is used there.)
 
  • #7
Thanks
 
Last edited:

1. What is thermal physics?

Thermal physics is a branch of physics that studies the behavior of systems in relation to temperature and heat. It involves the study of how energy is transferred between different forms of matter and how it affects the properties of materials.

2. What is the final temperature of a mixture?

The final temperature of a mixture is the temperature that the mixture reaches after two or more substances with different initial temperatures are combined. It is determined by the amount of heat energy transferred between the substances and their specific heat capacities.

3. How is the final temperature of a mixture calculated?

The final temperature of a mixture can be calculated using the principle of energy conservation, which states that the total energy of a closed system remains constant. The equation used is Q = m1c1ΔT1 + m2c2ΔT2, where Q is the total heat energy transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.

4. What factors affect the final temperature of a mixture?

The final temperature of a mixture is affected by several factors, including the initial temperatures of the substances, the mass and specific heat capacities of the substances, and the amount of heat energy transferred. The rate of heat transfer and any external influences, such as insulation, can also impact the final temperature.

5. What are some real-life applications of thermal physics and final temperature of mixtures?

Thermal physics and the calculation of final temperature of mixtures have many practical applications. Some examples include designing and improving energy-efficient buildings, understanding and predicting weather patterns, developing more efficient engines and refrigeration systems, and analyzing the thermal properties of materials for various industries such as construction, electronics, and manufacturing.

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