Thermal physics final temperature of mixture

  • #1
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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?
 

Answers and Replies

  • #2
Redbelly98
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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
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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
Redbelly98
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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
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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
Redbelly98
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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
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Thanks
 
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