What is the Final Temperature of the Mixture?

AI Thread Summary
The discussion revolves around calculating the final temperature of a mixture of two liquids with different temperatures and specific heat capacities. The user proposes using the equation cmΔT=cmΔT to find the final temperature, with variables defined for the higher and lower temperature mixtures. There is a question about the practicality of handling liquids with a density of 1070 kg/L, suggesting it would be very heavy. The conversation highlights the importance of understanding heat transfer principles in thermodynamics. The final temperature calculation remains unresolved in the discussion.
Brian13
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Homework Statement


Assuming no heat loss to the container or the environment, what is the final temperature (K) of a mixture of 9.28L of liquid 1 (density=1070kg/L & specific heat capacity of 4358J/kg/K) at 378K and 12.15L of a second liquid (density=1070kg/L & specific heat capacity of 1776J/kg/K) at 296K?

Homework Equations


I believe it should be cmΔT=cmΔT

The Attempt at a Solution


I came up with (cH)(mH)(TH-TM)=(cL)(mL)(TM-TL)
H= Mixture with higher temp
M=The middle temp we are looking for
L= Mixture with lower temp
 
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Brian13 said:

Homework Statement


Assuming no heat loss to the container or the environment, what is the final temperature (K) of a mixture of 9.28L of liquid 1 (density=1070kg/L & specific heat capacity of 4358J/kg/K) at 378K and 12.15L of a second liquid (density=1070kg/L & specific heat capacity of 1776J/kg/K) at 296K?

Homework Equations


I believe it should be cmΔT=cmΔT

The Attempt at a Solution


I came up with (cH)(mH)(TH-TM)=(cL)(mL)(TM-TL)
H= Mixture with higher temp
M=The middle temp we are looking for
L= Mixture with lower temp
Yeah, but what's the answer to the question?

Also, isn't a liquid with a density of 1070 kg/L rather heavy? You'd need a crane to pick up a cupful of this stuff.
 
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