Thermal physics final temperature of mixtureby AbsoluteZer0 Tags: final, mixture, physics, temperature, thermal 

#1
Feb213, 08:03 PM

P: 126

1. The problem statement, all variables and given/known data
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? 2. Relevant equations Conservation of Energy m_{c}CΔT_{c} = m_{h}CΔT_{h} 3. The attempt at a solution m_{c}CΔT_{c} = m_{h}CΔT_{h} (0.250)(390)(T_{f}  T_{i}) = (0.325)(4200)(T_{f}  T_{i}) (0.250)(390)(T_{f}  100) = (0.325)(4200)(T_{f  20}) Have I set up my equation properly? 



#2
Feb213, 08:55 PM

Mentor
P: 11,985

Let's back up to your starting equation, which really should say m_{c}C_{c}ΔT_{c} + m_{h}C_{h}ΔT_{h} = 0 (Conservation of energy means that the total change in energy is zero.) 



#3
Feb313, 01:23 PM

P: 126

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
Feb313, 09:07 PM

Mentor
P: 11,985

Thermal physics final temperature of mixtureAs 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_iT_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
Feb313, 10:16 PM

P: 126

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
Feb413, 05:25 PM

Mentor
P: 11,985

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
Feb413, 07:31 PM

P: 126

Thanks



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