Finding Initial Temperature Entropy and Heat Exchange

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SUMMARY

The discussion revolves around calculating the initial temperature of water when a 2.45 kg aluminum pan at 155°C is submerged in 3.58 kg of water, resulting in an entropy change of 162 J/K. The relevant equations include Q = mcΔT and ΔS = mc ln(T2/T1), with specific heat capacities of water (4184 J/kg*K) and aluminum (900 J/kg*K). Despite multiple attempts, the correct initial temperature of the water is confirmed to be 16°C, which can be derived by substituting values into the equations and solving for T_i.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer and entropy.
  • Familiarity with the equations Q = mcΔT and ΔS = mc ln(T2/T1).
  • Knowledge of specific heat capacities for water and aluminum.
  • Ability to solve logarithmic equations and systems of equations.
NEXT STEPS
  • Practice solving thermodynamic problems involving heat exchange and entropy changes.
  • Learn how to apply the first law of thermodynamics in practical scenarios.
  • Explore advanced topics in thermodynamics, such as reversible and irreversible processes.
  • Investigate the effects of varying mass and specific heat on temperature changes in mixed systems.
USEFUL FOR

Students studying thermodynamics, physics educators, and anyone involved in heat transfer calculations in engineering or scientific research.

Luke Strand
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Homework Statement


A 2.45 kg aluminium pan at 155 C is plunged into 3.58 kg of water. If the entropy change of the system is 162 J/k, what is the initial temperature of the water?

Homework Equations


Q = mcΔT ΔS=mcln(T_2/T_1) Q_water + Q_Aluminium = 0
c water = 4184 J/kg*K c aluminium = 900 J/kgK
3. Attempted Solutions
T_i Aluminum = 155
T_f is the same for the water and aluminium
T_i where T_i is the initial temperature of the water.
162 = (3.58)(4184)(ln(T_f/T_i) + (2.45)(900)ln((T-f)/(428))
(3.54)(4814)(T_f-T_i) + (2.45)(900)(T_f - 428)= 0

After solving this system of equations multiple times I can't seem to get the correct answer. The answer is 16 degrees C but I am not sure how to get there.
 
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Luke Strand said:

Homework Statement


A 2.45 kg aluminium pan at 155 C is plunged into 3.58 kg of water. If the entropy change of the system is 162 J/k, what is the initial temperature of the water?

Homework Equations


Q = mcΔT ΔS=mcln(T_2/T_1) Q_water + Q_Aluminium = 0
c water = 4184 J/kg*K c aluminium = 900 J/kgK
3. Attempted Solutions
T_i Aluminum = 155
T_f is the same for the water and aluminium
T_i where T_i is the initial temperature of the water.
162 = (3.58)(4184)(ln(T_f/T_i) + (2.45)(900)ln((T-f)/(428))
(3.54)(4814)(T_f-T_i) + (2.45)(900)(T_f - 428)= 0

After solving this system of equations multiple times I can't seem to get the correct answer. The answer is 16 degrees C but I am not sure how to get there.
Your setup is correct. Please show us what you did in attempting to solve the equations. Did you try substituting 16 C into the equations?
 

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