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How much energy is required to heat 36.0 g H2O from a liquid at 65°C to a gas at 115°C? The following physical data may be useful.ΔHvap = 40.7 kJ/molCliq = 4.18 J/g°CCgas = 2.01 J/g°CCsol = 2.09 J/g°CTmelting = 0°CTboiling = 100°C

Respuesta :

Answer:

energy required=qnet=87.75kJ

Explanation:

we will do it in three seperate step and then add up those value.

first step is to heat the sample of water upto 100C i.e upto boiling pont. because just after this sample of water started vaporization.

q 1= m c (T2-T1)

q1 = 36.0 g (4.18 J/gC) (100 - 65 C)

q1 = 5267 J =5.267kJ

next is to vaporize the sample at 100C

q2 = 36.0 g / 18.0 g/mol X 40.7 kJ/mol

q2= 81.4 kJ

Finally, heat the steam upto 115C

q3 = m c (T2-T1)

q 3= 36.0 g (2.01 J/gC)(115-100C)

q3 = 1085 J =1.085kJ

qnet=q1 +q2 +q3

energy required=qnet=87.75kJ

Cricetus

The required energy will be "87752 J".

According to the question,

  • ΔHvap = 40.7 kJ/mol
  • Liquid = 4.18 J/g°C
  • Gas = 2.01 J/g°C
  • Solid = 2.09 J/g°C
  • Melting temperature = 0°C
  • Boiling temperature = 100°C

Now,

The energy will be:

→ [tex]Q_1 = 36.0\times 3.18(100-65)[/tex]

        [tex]= 5267 \ J[/tex]

Moles of water will be:

= [tex]\frac{36.0}{18}[/tex]

= [tex]2[/tex]

then,

→ [tex]Q_2 = 2\times 40.7[/tex]

        [tex]= 81.4 \ kJ[/tex]

        [tex]= 81400 \ J[/tex]

→ [tex]Q_3 = 36.0\times 2.01(115-100)[/tex]

        [tex]= 1085 \ J[/tex]

hence,

The required energy will be:

= [tex]Q_1+Q_2+Q_3[/tex]

= [tex]5267+81400+1085[/tex]

= [tex]87752 \ J[/tex]

Thus the above answer is right.

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