Heat Transferred to 8,000-kg Aluminum Flagpole

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SUMMARY

The discussion centers on calculating the heat transferred to an 8,000-kg aluminum flagpole heated from 10°C to 20°C. Using the specific heat of aluminum at 0.215 cal/g°C, the heat transfer equation Q=mcΔT was applied. The calculation resulted in 17,200,000 cal, which converts to approximately 72,000,000 J. The consensus confirms the accuracy of this calculation.

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  • Understanding of specific heat capacity
  • Familiarity with the heat transfer equation Q=mcΔT
  • Knowledge of unit conversion between calories and joules
  • Basic principles of thermodynamics
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Homework Statement



An 8 000-kg aluminum flagpole 100-m long is heated by the sun from a temperature of 10°C to 20°C. Find the heat transferred (in J) to the aluminum if the specific heat of aluminum is 0.215 cal/g °C.

Homework Equations



Q=mc delta T

The Attempt at a Solution



(8000000g)(.215cal/g)(10)

so i get 17200000 cal convert to J
7.2e7 J

is that correct??
 
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i think so
 
Looks good to me too.
 

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