Answer:
5118.50 J
Explanation:
pΔv=nRΔT ;
therefore, ΔT=PV/nR
ΔT = (6.4×10^5)(3.2×10^(-3)/1×8.314
ΔT= 2.4633×10^2 = 246.33 K
specific heat at constant pressure is given as:
c_p = 3/2R
c_p = 12.5 J/mol K
Now, substitute in equation (1)
we know that
Q=ΔU+W ;
and
W=pΔV= 6.4×10^5×3.2×10^(-3) = 2048 J
now
ΔU=CvΔT = 12.465×246.33 =3070.50 J ;
therefore
Q=3070.50+2048= 5118.50 J
Answer:
The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.
Explanation:
A molar heat of vaporization of 40.66 kJ / mol means that 40.66 kJ of heat needs to be supplied to boil 1 mol of water at its normal boiling point.
To know the amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure, the number of moles represented by 3.11 g of water is necessary. Being:
the molar mass of water is:
H₂O= 2* 1 g/mole + 16 g/mole= 18 g/mole
So: if 18 grams of water are contained in 1 mole, 3.11 grams of water in how many moles are present?

moles of water= 0.1728
Finally, the following rule of three can be applied: if to boil 1 mole of water at its boiling point it is necessary to supply 40.66 kJ of heat, to boil 0.1728 moles of water, how much heat is necessary to supply?

heat= 7.026 kJ
<u><em>The amount of heat that is absorbed when 3.11 g of water boils at atmospheric pressure is 7.026 kJ.</em></u>
The curved top of a liquid is the meniscus