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18.1 g of water (initially at 20 oC) is poured onto 270 g of hot, aluminum metal (initially at 500 oC). Once all of the water has vaporized (and no heat is lost to the surrounding air) what will be the final temperature of the aluminum?

a.MPAl = 660 oC

b.Csolid Al = 24.2 J/mol oC

c.Cliquid Al = 29.3 J/mol oC

d.Cliquid water = 75.3 J/mol oC

e.Cwater vapor = 33.1 J/mol oC

f.ΔHfus water = 6.02 kJ/mol

g.ΔHvap water = 40.67 kJ/mol

Respuesta :

Answer:

The final temperature of the aluminum is 306°C

Explanation:

Have the follow:

The amount of heat is equal to:

[tex]q=n_{H_{2}O } (100-20)Cx_{H_{2}O } +n_{H_{2}O } H_{vapor} \\q=18.1g*\frac{1mol}{18.02g} *80C*\frac{75.3J}{molC} +18.1g*\frac{1mol}{18.02g} *\frac{40.67kJ}{mol} *\frac{1000J}{1kJ} =46901.3J[/tex]

Heat absorbed by the water = heat lost by aluminum

-q = qal

[tex]-q=n_{Al} C_{Al} (T-T_{Al} \\-46901.3J=270g*\frac{1mol}{26.98g} *\frac{24.2J}{molC} (T-500)\\T=306C[/tex]

onyebuchinnaji

The final temperature of the aluminum is 307 ⁰C.

The given parameters;

  • mass of the water, = 18.1 g
  • mass of the hot aluminum, = 270 g
  • initial temperature of the aluminum = 500 ⁰C
  • initial temperature of the water, = 20 ⁰C

The number of moles of the given water and aluminum is calculated as follows;

[tex]n_{H_2O} = \frac{18.1}{18} = 1 \ mole[/tex]

[tex]n_{Al} = \frac{270}{27} = 10 \ moles[/tex]

The heat required by water to vaporize is calculated as follows;

[tex]Q = nC\Delta t \ + \ nH_{vap}\\\\Q = (1\times 75.3)(100-20) \ + \ (1)(40,670)\\\\Q = 46,694 \ J[/tex]

Apply the principle of conservation of energy to determine the final of the aluminum;

Heat lost by aluminum = heat gained by water

[tex]nC\Delta _A_l = Q_w\\\\10\times 24.2 (500 - T) = 46,694\\\\121,000 - 242T = 46,694\\\\242T = 74,306\\\\T = \frac{74,306}{242} \\\\T = 307 \ ^0C[/tex]

Thus, the final temperature of the aluminum is 307 ⁰C.

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