Heat Capacity energy exchange problem

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SUMMARY

The final temperature of the water after being poured into a porcelain mug is calculated to be 94°C. The analysis uses the principle of energy conservation, where the heat lost by the water equals the heat gained by the mug. The specific heat capacities used are 4180 J/kg°C for water and 107 J/kg°C for porcelain. The calculations confirm that the setup and equations applied are correct, leading to the conclusion that the final temperature is indeed 94°C.

PREREQUISITES
  • Understanding of heat transfer principles, specifically Q=mcΔt.
  • Familiarity with specific heat capacities of materials.
  • Basic algebra for solving equations.
  • Knowledge of energy conservation in thermodynamics.
NEXT STEPS
  • Explore the concept of thermal equilibrium in heat exchange problems.
  • Learn about the specific heat capacities of different materials.
  • Study more complex heat transfer scenarios involving multiple substances.
  • Investigate the effects of varying mass and temperature on final equilibrium temperature.
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Students studying thermodynamics, physics educators, and anyone interested in understanding heat transfer calculations in practical scenarios.

sww_world
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Homework Statement
Mr. Gryphon was taking a break between classes and decided to make himself a cup of tea. He poured 250g of water at 95°C into an 85g porcelain mug which had an initial temperature of 19°C. What was the final temperature of the water for his tea? ( Not technically part of the question but the given heat capacities of water from a previous question was 4180 J/kg °C and porcelain was 107.)
Relevant Equations
Q=mc▲t, Qreleased + Qabsorbed = 0, T=Tf-Ti
Givens for water: m: 0.250kg of water
TW : 95°C
C=4180

Givens for mug: m=0.085kg
TM : 19°
c=107

Required: final temperature of water

Analysis/Solution: Qreleased+Qabsorbed=0, q=mc▲t

mw*cw*Tw + mp*cp*Tp = 0

(0.250)(4180) (T2-95) + (0.085)(107)(t2-19)=0

1045(t2-95) + 9.095(t2-19)=0

1045T2-99275 + 9.095T2-172.805=0

1054.095T2 - 99447.805 = 0

99447.805/1054.095 = t2

t2=94

Statement: The final temperature of the water for his tea was 94 degreesI'm not sure here if what I did with the temperature was correct, or if I made a mistake somewhere else. (I would also appreciate if you guys tell me if my formatting is acceptable/too hard too look at?)
 
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sww_world said:
Problem Statement: Mr. Gryphon was taking a break between classes and decided to make himself a cup of tea. He poured 250g of water at 95°C into an 85g porcelain mug which had an initial temperature of 19°C. What was the final temperature of the water for his tea? ( Not technically part of the question but the given heat capacities of water from a previous question was 4180 J/kg °C and porcelain was 107.)
Relevant Equations: Q=mc▲t, Qreleased + Qabsorbed = 0, T=Tf-Ti

Givens for water: m: 0.250kg of water
TW : 95°C
C=4180

Givens for mug: m=0.085kg
TM : 19°
c=107

Required: final temperature of water

Analysis/Solution: Qreleased+Qabsorbed=0, q=mc▲t

mw*cw*Tw + mp*cp*Tp = 0

(0.250)(4180) (T2-95) + (0.085)(107)(t2-19)=0

1045(t2-95) + 9.095(t2-19)=0

1045T2-99275 + 9.095T2-172.805=0

1054.095T2 - 99447.805 = 0

99447.805/1054.095 = t2

t2=94

Statement: The final temperature of the water for his tea was 94 degreesI'm not sure here if what I did with the temperature was correct, or if I made a mistake somewhere else. (I would also appreciate if you guys tell me if my formatting is acceptable/too hard too look at?)
All good.
 
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