Thermodynamics - Change in Temp

AI Thread Summary
To calculate the temperature change of a hiker's body during a 2.9-hour hike, it's essential to consider the total energy used over that time. The hiker expends 191 kcal/hr, which converts to 222 J/s, leading to a total energy of approximately 2,314,560 J over 2.9 hours. Since 80% of this energy converts to heat, the heat flow into the body is about 1,851,648 J. The specific heat capacity of the human body is 3475.04 J/kg/K, and using the formula Q = mc(change in temperature), the correct calculation yields a temperature change of approximately 1.54 K, not the initially calculated 8.29 x 10^-4 K. Accurate energy conversion and time consideration are crucial for the correct result.
Jess_18033152
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Homework Statement


A 77.1 kg hiker uses 191 kcal hr-1 (converted to 222.133 J.s ?) 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 2.9 hour (converted to 10440s) 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. (converted to 3475.04 J/kg/K ?)

The answer needs to be given in Kelvin,

Homework Equations


Q = mc(change in temperature)

The Attempt at a Solution


Change in temp = Q/mc = 222.133/(77.1 x 3475.04)
=8.29 x 10^-4 K
Answer is wrong, I feel like I need to be using the time given also...
 
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Jess_18033152 said:

Homework Statement


A 77.1 kg hiker uses 191 kcal hr-1 (converted to 222.133 J.s ?) 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 2.9 hour (converted to 10440s) 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. (converted to 3475.04 J/kg/K ?)

The answer needs to be given in Kelvin,

Homework Equations


Q = mc(change in temperature)

The Attempt at a Solution


Change in temp = Q/mc = 222.133/(77.1 x 3475.04)
=8.29 x 10^-4 K
Answer is wrong, I feel like I need to be using the time given also...
Q is the total heat flow generated by the hiker over the 2.9 hour hike. How much energy does he use in 2.9 hours? How much of that results in heat flow into the body?

AM
 
Last edited:
Hi jess, you must be in Kevin’s class! https://www.physicsforums.com/threads/heat-question.945804/

Indeed, you need to use the time. Note that 191 kcal/hr is 222 J/s (not J.s)

Q is supposed to be a quantity of energy but for Q you’ve plugged in 222 J/s which is a measure of power not energy.
 
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