Chemistry question on vaporisation

Topps

An average man weights 70kg and produces 10,460KJ of heat per day.

(i) Suppose that the man is an isolated system and that his heat capacity is 4.18JK-1g-1 ;if his temperature were 37 degrees Celsius at a given time, what would his temperature be 24 hours later?

OK I got this part right at 72.7 degrees Celsius but part two is the part I can figure out.

(ii) The man is in fact an open system, and the main mechanism for maintaining his temperature constant is by the evaporation of water from his body. If the enthalpy of vaporization of water at 37 degrees Celsius is 43.4KJ mol-1, how many grams of water need to be evaporated per day (i.e 24 h) to keep his temperature constant?
Given: GMM for H2O = 18 g mol-1

UnholyGregor

COuld very well be wrong here, but I'm getting an answer of 4318.3g of water, which seems like an awful lot...

Here's how I got that:

man makes 10460 kj energy/day
1mol water evaporating=43.4kj
So, 18g water removes 43.4jk
So, 1g water= 2.4kj

So, 10460/2.4 = no. grams of water needed to completely remove all the energy the man produces.

=4.318kg water.

Seems like a huge oversimplification, and Ill probably facepalm when i realize how wrong it is...

Topps

Thank for that. I would of been concerned by the figure but by the temperature increase in part 1 anything is possible for this part:D

UnholyGregor

Aye, seems like one of those deceptively easy questions, like you get the answer and are like "that cant be all, there has to be some trick here", getting all paranoid lol.

And you're welcome.. anything to avoid studying real chemistry...

Topps

192g dm-3 to SI anyone?:(

MathsTeacherD

Topps wrote: »

192g dm-3 to SI anyone?:(

Not sure where this question is coming from, but...
1 dm ('1 decimetre') = 10cm and 1 dm^3 = 1 litre:
1 dm^3 = 1 x (10cm)^3 = 10cm x 10cm x 10cm = 1000cm^3 = 1 litre

also 1 m^3 = 1 x (100cm)^3 = 100cm x 100cm x 100cm = 1000000 cm^3
so 1 m^3 = 1000 litres (L)

so 192 g dm^-3 = 192 g L^-1 = 192 kg m^-3