Suppose that 0.50 g of water at 25 ∘c condenses on the surface of a 51-g block of aluminum that is initially at 25 ∘c. if the heat released during condensation goes only toward heating the metal, what is the final temperature (in ∘c) of the metal block? (the specific heat capacity of aluminum is 0.903 j/g∘c and the heat of vaporization of water at 25 ∘c is 44.0 kj/mol.)

Moles water = 0.50 g / 18.02 g/mol=0.0277

heat lost by water = 0.0277 mol x 44.0 kJ/mol=1.22 kJ => 1220 J
heat gained by aluminum = 1220 J

1220 = 51 x 0.903 ( T - 25) = 46.05 T - 1151
1220 + 1151 = 46.05T

46.05T = 2371
T = 51.48 °C (51 °C at two significant figures)

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johanrusli johanrusli · Answer 2 · 2019-12-28T10:12:40+01:00

The final temperature of the metal block is about 51.5°C

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Further explanation

Let's recall the Specific Heat Capacity formula as follows:

[tex]\boxed {Q = m c \Delta t }[/tex]

where :

Q = heat energy ( J )

m = mass of object ( kg )

c = specific heat capacity ( J/kg°C )

Δt = change in temperature ( °C )

Let us now tackle the problem!

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Given:

mass of water = m = 0.50 g

mass of aluminium = M = 51 g

initial temperature of aluminium = to = 25°C

specific heat capacity of aluminium = c = 0.903 J/g°C

heat of vaporization of water = Lv = 44.0 kJ/mol =44000 J/mol

molar mass of water = 18 g/mol

Asked:

final temperature of aluminium = t = ?

Solution:

We will use Conservation of Energy formula as follows:

[tex]\texttt{Heat Released by Water } = \texttt{ Heat Gained by Aluminium }[/tex]

[tex]n L_v = M c \Delta t[/tex]

[tex]( m \div \texttt{Molar Mass} ) L_v = M c ( t - t_o )[/tex]

[tex]( 0.50 \div 18 ) \times 44000 = 51 \times 0.903 \times ( t - 25 )[/tex]

[tex]2373.55 = 46.053 t[/tex]

[tex]t \approx 51.5^oC[/tex]

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Answer details

Grade: High School

Subject: Mathematics

Chapter: Energy