April 24, 2024, 11:41:53 AM
Forum Rules: Read This Before Posting


Topic: Water evaporation question  (Read 3394 times)

0 Members and 1 Guest are viewing this topic.

Offline JobNick

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Water evaporation question
« on: June 21, 2014, 03:54:57 AM »
Hey all,
I don't know how to solve this one:
Human body produces 150 Kj of heat per Kg of body weight through metabolic processes. Human body is open system, and the main mechanism for loosing heat is water vapor. Whats is the water volume (in liters) that has to be vapored every day to keep the same body temperature of a parson whose weight is 76 Kg. Hv = 40.65 Kj (Enthalpy of vaporization) and water density is 1 g/mol.

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3652
  • Mole Snacks: +222/-42
  • Gender: Male
Re: Water evaporation question
« Reply #1 on: June 21, 2014, 04:40:43 AM »
Hey all,
I don't know how to solve this one:
Human body produces 150 kJ of heat per Kg of body weight through metabolic processes. Human body is open system, and the main mechanism for loosing heat is water vapor. Whats is the water volume (in liters) that has to be vapored every day to keep the same body temperature of a parson whose weight is 76 kg. Hv = 40.65 kJ (Enthalpy of vaporization) and water density is 1 g/mol.

1) How much heat does this body produce?

Offline JobNick

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Water evaporation question
« Reply #2 on: June 21, 2014, 06:05:39 AM »
150 Kj/Kg * 76 Kg = 11400 Kj (kilojoules)

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3652
  • Mole Snacks: +222/-42
  • Gender: Male
Re: Water evaporation question
« Reply #3 on: June 21, 2014, 06:58:59 AM »
150 Kj/Kg * 76 Kg = 11400 Kj (kilojoules)

What are the units of your enthalpy of vaporization?

Offline JobNick

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Water evaporation question
« Reply #4 on: June 21, 2014, 10:06:34 AM »
Kj = Kilograms, Kj = kilojoules.
So it would be:
Person weight: 76 kilograms
Water Enthalpy of vaporization = 40.65 kilojoules
Human body heat production: 150 kilojoulesof heat per kilograms
water density is: 1 gram/mol

Offline sjb

  • Global Moderator
  • Sr. Member
  • ***
  • Posts: 3652
  • Mole Snacks: +222/-42
  • Gender: Male
Re: Water evaporation question
« Reply #5 on: June 21, 2014, 12:38:54 PM »
Kj = Kilograms, Kj = kilojoules.
So it would be:
Person weight: 76 kilograms
Water Enthalpy of vaporization = 40.65 kilojoules
Human body heat production: 150 kilojoulesof heat per kilograms
water density is: 1 gram/mol

Not quite sure what it is you're doing here, can you explain?

I was trying to get you to convert the 11400 kJ given off by metabolic processes into some amount of water vaporised.

Offline JobNick

  • New Member
  • **
  • Posts: 4
  • Mole Snacks: +0/-0
Re: Water evaporation question
« Reply #6 on: June 21, 2014, 02:10:11 PM »
Actually I didn't do anything. I've just wrote the questions data in rows.

Offline Furanone

  • Chemist
  • Full Member
  • *
  • Posts: 211
  • Mole Snacks: +34/-2
  • Gender: Male
  • Actually more a Food Chemist
Re: Water evaporation question
« Reply #7 on: July 22, 2014, 05:57:07 PM »
I was with you both until I wasn't
"The true worth of an experimenter consists in pursuing not only what he seeks in his experiment, but also what he did not seek."

--Sir William Bragg (1862 - 1942)

Sponsored Links