What is the Final Temperature of the Mixture?

Click For Summary
SUMMARY

The final temperature of a mixture of two liquids can be calculated using the principle of conservation of energy, specifically the equation cmΔT = cmΔT. In this case, the first liquid has a density of 1070 kg/L and a specific heat capacity of 4358 J/kg/K at an initial temperature of 378 K, while the second liquid has the same density but a specific heat capacity of 1776 J/kg/K at 296 K. By applying the equation and solving for the final temperature, the resulting temperature of the mixture can be determined without any heat loss to the environment.

PREREQUISITES
  • Understanding of specific heat capacity
  • Knowledge of the conservation of energy principle
  • Ability to perform calculations involving density and volume
  • Familiarity with temperature scales (Kelvin)
NEXT STEPS
  • Study the concept of heat transfer in mixtures
  • Learn about the calculation of final temperature in thermodynamic systems
  • Explore the implications of specific heat capacity in different materials
  • Investigate the effects of density on liquid mixtures
USEFUL FOR

Chemistry students, physics students, and anyone interested in thermodynamics and heat transfer principles will benefit from this discussion.

Brian13
Messages
4
Reaction score
0

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
 
Physics news on Phys.org
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.
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
961
Replies
12
Views
1K
Replies
6
Views
1K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 5 ·
Replies
5
Views
5K