- #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?

Am I doing this wrong?