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klasskru [66]
1 year ago
15

A goldsmith melts 12.4 grams of gold to make a ring. The temperature of the gold rises from 26°C to 1064°C, and then the gold me

lts completely. If gold’s specific heat is 0.1291 joules/gram degree Celsius and its heat of fusion is 63.5 joules/gram, how much energy is gained by the gold?
Chemistry
1 answer:
ratelena [41]1 year ago
5 0

Answer: 2,448 J


Explanation:


1) The process in which the 12.4 g gold sample was melted can be splited in two differentiated stages: i) the increase of temperature from 26°C to 1064°C, and ii) the melting perse, which is at constant temperature.


Each stage is ruled by it specific thermodynamic model, which you can use to compute how much energy the gold gained.


2) Rising temperature


The heat energy gained by the gold to increase the temperature is given by the equation Q₁ = mCΔT.


Where m is the mass of gold, 12.4g, C is the specific heat, 0.1291 J/g°C, and ΔT is the change of temperature, 1,064°C - 26°C. Hence:

Q = 12.4g × 0.1291 J/g°C × (1,064°C - 26°C) = 1,661 J.


3) Melting


The heat energy during this stage is computed with the formula for the latent heat of fusion:

Q₂ = m × latent heat of fusion = 12.4  g × 63.5 J/g = 787.4 J ≈ 787 J.


3) The total heat energy gained, Q, is the sum of the energy gained during the two stages:

Q = Q₁ + Q₂ = 1,661 J + 787 J = 2,448 J ← answer

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Now put all the given values in the above formula, we get:

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The formula used for \Delta G_{rxn} is:

The given reaction is:

A(aq)\rightleftharpoons B(aq)

\Delta G_{rxn}=\Delta G^o+RT\ln Q

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R = gas constant = 8.314\times 10^{-3}kJ/mole.K

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[A] = concentration of A = 1.8 M

[B] = concentration of B = 0.55 M

Now put all the given values in the above formula 1, we get:

\Delta G_{rxn}=(-5980J/mol)+[(8.314J/mole.K)\times (310K)\times \ln (\frac{0.55}{1.8})

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4 0
1 year ago
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MA_775_DIABLO [31]
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