How do I rearrange this equation to solve for CCu? And if you could show me how to solve it with the answer that'd be super helpful!  Ti = 26C Tf = 18.9C Cw = 1.00 cal/gram celcius  Cc = 0.200 cal/gram celcius  MC 25.2g MCU 161.6g M­C + MW  71.6g MW =71.6-25.2 46.4g

University Physics Volume 2
18th Edition
ISBN:9781938168161
Author:OpenStax
Publisher:OpenStax
Chapter3: The First Law Of Thermodynamics
Section: Chapter Questions
Problem 66P: One of a dilute diatomic gas occupying a volume of 10.00 L expands against a constant pressure of...
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How do I rearrange this equation to solve for CCu?

And if you could show me how to solve it with the answer that'd be super helpful! 

Ti = 26C

Tf = 18.9C

Cw = 1.00 cal/gram celcius 

Cc = 0.200 cal/gram celcius 

MC

25.2g

MCU

161.6g

C + MW 

71.6g

MW =71.6-25.2

46.4g

Q; = Q, + Q: - caMa(T;– 0°C) = c¸M„(T, – T;) + c_Mc(T, – T;)
Transcribed Image Text:Q; = Q, + Q: - caMa(T;– 0°C) = c¸M„(T, – T;) + c_Mc(T, – T;)
The specific heat for the copper sample is calculated as follows:
Heat lost by water
Q: = c,M«(T, – T)
Heat lost by the calorimeter cup
Q: = cMc(T, – T)
Heat gained by the metal sample Q = coMa(T; – 0°C)
Q = Qi + Q; = coM(T,- 0°C) = c,Mw(T, – T) + cM(T, – T,)
(T1-4)
The only unknown in the above equation is car
Transcribed Image Text:The specific heat for the copper sample is calculated as follows: Heat lost by water Q: = c,M«(T, – T) Heat lost by the calorimeter cup Q: = cMc(T, – T) Heat gained by the metal sample Q = coMa(T; – 0°C) Q = Qi + Q; = coM(T,- 0°C) = c,Mw(T, – T) + cM(T, – T,) (T1-4) The only unknown in the above equation is car
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