Solve Work/Energy Problem: 5.9E4 s Answer

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SUMMARY

The discussion centers on solving a heat transfer problem involving a cubical cooler with specific dimensions and thermal properties. The correct answer for the time it takes for water to warm from 2 degrees C to 3 degrees C is established as 5.9E4 seconds. Key equations utilized include Q/T = KA(T2-T1)/L and Q = cm(delta)T, with emphasis on calculating the surface area of the cooler and the mass of water to determine the heat transfer accurately. Participants highlight the importance of unit consistency in calculations to achieve correct results.

PREREQUISITES
  • Understanding of heat transfer principles, specifically conduction.
  • Familiarity with the concept of R-value in thermal insulation.
  • Knowledge of specific heat capacity, particularly for water (4190 J/kg K).
  • Ability to perform calculations involving surface area and volume of geometric shapes.
NEXT STEPS
  • Study the derivation and application of the heat transfer equation Q/T = KA(T2-T1)/L.
  • Learn how to calculate the surface area and volume of a cube for thermal analysis.
  • Explore the concept of thermal resistance and its impact on heat transfer rates.
  • Investigate practical applications of heat transfer in real-world scenarios, such as refrigeration and insulation.
USEFUL FOR

Students in physics or engineering disciplines, particularly those focusing on thermodynamics and heat transfer, as well as educators looking to enhance their teaching materials on thermal dynamics.

jtb93
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Homework Statement


A cubical cooler, 0.45 m on a side, has an R-value of 3.2 m2 K/W. The sides of the cooler are 0.030 m thick. The cooler is filled with water at a temperature of 2 degrees C and is placed in a room with a temperature of 19 degrees C. The heat capacity of water is 4190 J/kg K. How long does it take for the water to warm to 3 degrees C?

Homework Equations


Q/T=KA(T2-T1)/L
Q=cm(delta)T

The Attempt at a Solution


Q/T=(0.14)(0.453)(19-2)/0.45
also 6.5 W(4190)(19-2)m[/SUP][/SUP]

Correct answer is 5.9E4 s. My attempts got nothing close to that.

[mentor's note: edited to fix formatting]
 
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jtb93 said:

Homework Statement


A cubical cooler, 0.45 m on a side, has an R-value of 3.2 m2 K/W. The sides of the cooler are 0.030 m thick. The cooler is filled with water at a temperature of 2 degrees C and is placed in a room with a temperature of 19 degrees C. The heat capacity of water is 4190 J/kg K. How long does it take for the water to warm to 3 degrees C?

Homework Equations


Q/T=KA(T2-T1)/L
Q=cm(delta)T

The Attempt at a Solution


Q/T=(0.14)(0.453)(19-2)/0.45
also 6.5 W(4190)(19-2)m[/SUP][/SUP]

Correct answer is 5.9E4 s. My attempts got nothing close to that.

Use the Preview Button to check the formatting of your thread before you post.

As far as your calculations are concerned, what the heck are you doing? You haven't even figured out the surface area of the cooler through which heat is going to enter from the surroundings. You haven't even calculated the amount of water in the cooler, let alone the amount of heat it takes to warm the contents 1° C. :frown:

If nothing else, plug the units of the various quantities into your calculations to see if you are computing the right results. :wink:
 

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