Respuesta :
Answer:
1918.62 kj
Explanation:
The heat of vaporization of water is 540cal/g.
Temperature of vaporization of water is 100°c.
Mass of water is 8.5grams.
Heat released = mc(tita)
Heat released= 8.5*540*100
Heat released = 459000 cal
1 kilocalorie =
4.184 kilojoules
459000 cal = 459 kc
459 kc = 459*4.18
459 kc = 1918.62 kj
Answer:
Heat Loss of, q = 19.21 KJ
Explanation:
Given:-
- The mass of water, m = 8.5 g
- The latent heat of vaporization of water, Lv = 2,260 kJ/kg.
Find:-
how much heat will be released from the body when the sweat evaporates?
Solution:-
- The heat loss of water assuming conditons of evaporation only depedns on the amount ( mass ) of water and its latent heat of vaporisation i.e ( its phase change from liquid to gaseous form ). We can write the heat loss (q) as following:
q = m*Lv
- Plug in the given values:
q = ( 8.5 / 1000 ) * ( 2,260 )
q = 19.21 KJ
