- #1
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 celcius
273.15K+4.066=277.216K
277K (3sf)
Am I doing this wrong?