What mass of water must evaporate from the skin of a 70.0 kg man to cool his body 1.00 ∘c? the heat of vaporization of water at body temperature (37.0 ∘c) is 2.42×106j/(kg⋅k). the specific heat of a typical human body is 3480 j/(kg⋅k) . what volume of water must the man drink to replenish the evaporated water?

100 cc

Explanation:

Heat released in cooling human body by t degree

= mass of the body x specific heat of the body x t

Substituting the data given

Heat released by the body

= 70 x 3480 x 1

= 243600 J

Mass of water to be evaporated

= 243600 / latent heat of vaporization of water

= 243600 / 2420000

= .1 kg

= 100 g

volume of water

= mass / density

= 100 / 1

100 cc

1 / 10 litres.

  1.the balanced   chemical equation is as below

cu +   2 ag(no)₃ → cu(no₃)₂   +2 ag

2.the   much silver   produced   is   50.9 g

step 1: find the   moles of cu

moles =   mass÷   molar mass

  from periodic table the molar mass of cu = 63.5 g/mol

moles is therefore = 15.00 g÷ 63.5 g/mol = 0.236   moles

step 2: use the mole ratio to determine the   moles of ag

from equation above   cu: ag   is 1: 2

therefore the moles of ag   =0.236   moles   x 2/1 =0.472   moles

step 3: find the mass   of ag

mass = moles x molar mass

from periodic table   the molar mass of ag =107.87 g/mol

mass   = 0.472   moles x 107.87 g/mol = 50.9 g

3.   the   much   silver   collected worth is   1.797 oz

  convert   50.9 g   to oz

that is   1 gram = 0.0353 oz

          50.9 grams =?   oz

by cross multiplication

={(50.9 grams x 0.0353 oz) / 1   gram}   =1.797 oz