Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 14, Problem 25P
To determine

If a person can give off 185 kCal of heat in 25 minutes by evaporation of water (at 200 C)from the skin, then how much water gets lost.

Expert Solution & Answer
Check Mark

Answer to Problem 25P

Solution:

0.2984 kg water is lost by the skin.

Explanation of Solution

Given:

Q=185 kCalQ=185×106Cal

Formula used:

The heat energy absorbed or released to raise the temperature by one degree is given by;

Q=mcΔT

Here,

m = mass of the substance.

C = specific heat.

◻ T= Temperature difference.

Calculation:

Evaporation of sweating starts at 1000C.

ΔT=1000C200CΔT=800C 

Now, the heat needed to convert water (1000C)to steam (at 1000C)is given by;

Q2=mLQ2=m(2260 kJ/kg)

So, total heat required is given by;

Q=Q1+Q2Q=(334.88 kJ/kg +2260 kJ/kg)×m

Q=(2594.88kJ/kg)×m …(1)

As Q=185 kCalQ=185×103cal ×4.186 J/calQ=774.41×103 J      (2)

Equating (1)and (2);

Heat needed to raise the temperature at this level is given by;

(2594.88kJ/kg)×m=774.41 kJm=0.2984 kg

Chapter 14 Solutions

Physics: Principles with Applications

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