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the volume of each glass is 250 ml. how much heat, in kJ, does the body have to supply to maintain the temperature at 37 deg C?

2006-09-24 19:14:06 · 4 answers · asked by Anonymous in Science & Mathematics Chemistry

okay, i've gotten that far.. but then how much heat is required to raise the temperature of 800g of snow from 0 to 37 deg. C?

2006-09-24 19:24:05 · update #1

4 answers

Mass of water (from the Volume) times specific heat of water times temperature difference.

250 mL * 4 = 1000 mL = 1.000 L
(your teacher probably wants you to assume that this is the same as 1000 grams of water)

Water has a specific heat of 4.184 J/g-°C

The difference in temperature is 37 degrees - 3 degrees = 34 degrees Celcius.

1000 g * 4.184 J/g-°C * 34 °C = 142256 J

142256 J / 1000 J/kJ = 142.256 kJ

to 4 significant figures = 142.3 kJ

Taking into effect the density of water as 0.999992 g/mL at 3.0 °C would give the answer as 142255 J which would still be 142.3 kJ when rounded.

2006-09-24 19:20:08 · answer #1 · answered by Richard 7 · 70 0

I'm not sure what kJ are-kiloJoules? but it takes one calorie to increase one ml of water one degree C, so multiply the liter of water times the calorie change, 1000 x 34=3400 calories. Now convert calories to kJ, and you're there...

2006-09-24 19:19:29 · answer #2 · answered by lee m 5 · 0 1

q=mCDT
DT=37-3=34C
mw=dw*Vw=250ml*1g/ml=250g
Cw= 4.184J/g C
q=250g*4.184J/g C*34*(1KJ/1000J)=35.57 KJ

2006-09-24 19:23:51 · answer #3 · answered by Fabrizio G 2 · 0 0

I purely drink water chilly, as i do in comparison to coffee and intensely hardly drink tea. yet sometimes I sip a mug of soup! in any different case I frequently drink between 1L to a million.5 L an afternoon. I shop a available bottle of water of 1L, it is how i understand. yet I additionally drink different beverages which incorporate fruit juices, and so on.

2016-12-18 16:28:44 · answer #4 · answered by ? 4 · 0 0

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