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Topic: Energy consumption  (Read 7522 times)

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Offline BreakingBad20

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Energy consumption
« on: April 28, 2011, 10:51:36 AM »
(a)      An average man weighs about 70kg and produces about 10 460 kJ of heat per day. His heat capacity is
 4.18 J K-1 g -1
(ii) A man is in fact an open system, and the main mechanism for maintaining his temperature constant is evaporation of water.  If the enthalpy of vaporization of water at 37oC  is 43.4 kJ mol -1, how much water needs to be evaporated per day to keep his temperature constant?

I got part (i) which is irrelevant to this question. Here is my answer but I don't think it's right:
E = (Mman)(C)(deltaT) + (Mwater)(L)
10460000 = (70)(4.18)(37) + M(43400)
M = 240.76kg

Offline enahs

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Re: Energy consumption
« Reply #1 on: April 28, 2011, 11:15:19 AM »
Your vaporization enthalphy is in units of mols, not mass or liters. Also, your heat capacity for a man is in units of grams, but you used in your calculations kg.

Also, how does a man have a  :delta: T of 37? What is the units of your heat capacity? Does it have T in it?


Units!



Perhaps you should post the whole question, as part I might not be irrelevant and it might help.

Offline BreakingBad20

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Re: Energy consumption
« Reply #2 on: April 28, 2011, 11:21:56 AM »
Q.1   
(a)      An average man weighs about 70kg and produces about 10 460 kJ of heat per day.

(i) Suppose that a man is an isolated system and that his heat capacity is
 4.18 J K-1 g -1; if his temperature were 37C at a given time, what would his temperature be 24h later?

(ii) A man is in fact an open system, and the main mechanism for maintaining his temperature constant is evaporation of water.  If the enthalpy of vaporization of water at 37C  is 43.4 kJ mol -1, how much water needs to be evaporated per day to keep his temperature constant?    

So should I leave delta T to be zero because the temperature has to be constant leaving the equation to be used E = ML for the water?

Offline enahs

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Re: Energy consumption
« Reply #3 on: April 28, 2011, 11:22:56 AM »
I re-updated my post after you posted, go back and check it real quick. Units!

Offline BreakingBad20

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Re: Energy consumption
« Reply #4 on: April 28, 2011, 11:45:08 AM »
I understand what you mean about the units but my new problem is that if the temperature of the man is to be kept constant does that not mean delta T (either in kelvin or degrees Celsius) is going to be zero? This would then cancel out (Mman)(C)(deltaT) would it not?

Offline enahs

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Re: Energy consumption
« Reply #5 on: April 28, 2011, 12:42:16 PM »
The question says his temperature is constant. So essentially you are assuming there is this steady state of energy flowing in and out of the man, and I believe you can neglect his heat capacity.

Offline BreakingBad20

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Re: Energy consumption
« Reply #6 on: April 28, 2011, 01:22:03 PM »
So am I right to assume E = (Mwater)(L) for part (ii) ?
if so:
10460 kJ = (M)(43.4 kJ/mol)
Mwater = 241.01 mol
> mass in grams = moles x Mr
> mass in grams = 241.01 x 18.016
> mass in grams = 4342.03616g
Mass in Kilograms = 4.342 Kg
This certainly looks a lot more accurate than my first attempt. ;D

Offline enahs

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Re: Energy consumption
« Reply #7 on: April 28, 2011, 01:50:42 PM »
It is totaly possible to sweat 4-5 liters a day. I would say that is not average at all, for just body maintenance. Unless you live in a dessert, or workout a lot, etc.  So I feel happy about it I guess.

Offline BreakingBad20

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Re: Energy consumption
« Reply #8 on: April 28, 2011, 02:40:53 PM »
ya i suppose ... thanks anyway!

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