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

[Kevin Kim]

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?

 

[Dr Dr news]

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..