Specific Heat Problem (did something wrong?)

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SUMMARY

The discussion centers on a specific heat problem involving a 55.0 g mass of iron and 100 mL of water at an initial temperature of 20 degrees Celsius, reaching a final temperature of 42.7 degrees Celsius. The correct equation for heat transfer is established as 55 * 0.451 * (42.7 - X) = -100 * 4.184 * (42.7 - 20), where X represents the initial temperature of the iron. A common mistake identified was the omission of the negative sign in the heat transfer equation, leading to incorrect calculations. The correct application of the specific heat formula is crucial for accurate results in thermal energy problems.

PREREQUISITES
  • Understanding of specific heat capacity, particularly for iron (Cs = 0.451 J/gC) and water (Cs = 4.184 J/gC).
  • Familiarity with the concept of heat transfer and the principle of conservation of energy.
  • Ability to manipulate algebraic equations to solve for unknown variables.
  • Basic knowledge of units of measurement in physics, specifically grams and degrees Celsius.
NEXT STEPS
  • Review the principles of heat transfer and specific heat calculations in thermodynamics.
  • Practice solving similar problems involving heat exchange between different substances.
  • Explore the concept of calorimetry and its applications in measuring heat transfer.
  • Learn about the implications of energy conservation in closed systems.
USEFUL FOR

Students studying physics or chemistry, particularly those focusing on thermodynamics, as well as educators looking for examples of specific heat problems and solutions.

Lori

Homework Statement


Heated iron with mass of 55.0 g was added to 100 mL of water at 20 degree Celsius. Assuming no energy transfer to the surroundings and that the final temperature of the system is 42.7 degrees C, calculate initial temperature of the iron.

Mass of Iron = 55.0 g
mass of Water = 100g
initial Temp of Water = 20 degrees C
Final temp of Water/Fe = 42.7 degrees
Want: initial temp of Fe

Given Cs of Fe = 0.451 J/gC
Cs of Water = 4.184 J/gC

Homework Equations



heat of Iron = -heat of water (q=-q)

mCT = mCT

The Attempt at a Solution


mCT of Fe = -mCT of water
[/B]
plug in:
X= initial temp of Fe
55 * 0.451*(42.7-X) = 100*4.184*(42.7-20)

When i solve for X , i get a large negative number = -9497.68 which is clearly not the final temperature!

Was wondering where i went wrong!
 
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It looks like you lost the negative sign. It should be
$$55 \cdot 0.451\cdot(42.7-X) = -100\cdot 4.184\cdot (42.7-20)$$
 
NFuller said:
It looks like you lost the negative sign. It should be
$$55 \cdot 0.451\cdot(42.7-X) = -100\cdot 4.184\cdot (42.7-20)$$
Wow, didn't realize that. Thanks !
 

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