Heat Question -- Temperature rise of a hiker on a long hike

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SUMMARY

The discussion centers on calculating the temperature rise of an 80.4 kg hiker during a 1.6-hour hike, where the hiker expends 212 kcal hr-1 of energy. It is established that 80% of this energy converts to heat, resulting in a total heat energy of 1,135,912.96 J. Using the specific heat capacity of the human body at 0.83 kcal kg-1 °C-1, the temperature change is calculated to be approximately 4.066 °C, which translates to a final temperature of 277.216 K. The key takeaway is that the temperature change in Celsius is equivalent to that in Kelvin.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with the specific heat capacity concept.
  • Basic knowledge of energy conversion (calories to joules).
  • Proficiency in algebraic manipulation of equations.
NEXT STEPS
  • Study the concept of specific heat capacity in more detail.
  • Learn about energy conversion factors, particularly between calories and joules.
  • Explore thermodynamic equations related to heat transfer and temperature change.
  • Investigate the physiological effects of temperature changes on the human body during physical activity.
USEFUL FOR

This discussion is beneficial for students in physics or thermodynamics, hikers interested in understanding their body's heat management, and educators teaching concepts related to energy and heat transfer.

Kevin Kim
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Homework Statement


A 80.4 kg hiker uses 212 kcal hr-1 (3 s.f.) of energy whilst hiking. Assuming that 20% of this energy goes into useful work and the other 80% is converted to heat within the body, calculate the temperature change, in units of Kelvin (K), of the hiker's body during a 1.6 hour long hike.

Assume that none of this generated heat is transferred to the environment during the hike. The average specific heat capacity of a human body is 0.83 kcal kg-1 oC-1.

Homework Equations


Q=mcDeltaT
1 Cal= 4.186 J

The Attempt at a Solution



m=80.4kg Q= 212000cal*4.186J*1.6hours*0.8 (converted to heat within body)= 1135912.96J c= 830*4.186= 3474.38
[/B]
Q/mc= DeltaT
1135912.96J/80.4kg*3474.38= 4.066 celsius

273.15K+4.066=277.216K
277K (3sf)

Am I doing this wrong?
 
Physics news on Phys.org
Mr. Kim: The question is What is the temperature rise? That was your first calculation and it was correct. You should note that ΔT on the Celsius scale is the same as ΔT on the Kelvin scale..
 

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