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sweet-ann [11.9K]
3 years ago
6

A 60.0 g block of iron that has an initial temperature of 250. °C and 60.0 g bloc of gold that has an initial temperature of 45.

0 °C are brought into contact with one another. Assuming no heat is lost to the surroundings, what will be the temperature when the two metals reach thermal equilibrium? The specific heat capacity of iron = 0.449 J/g•°C and gold = 0.128 J/g•°C.
Chemistry
1 answer:
Maslowich3 years ago
5 0

Answer:

The final temperature at the equilibrium is 204.6 °C

Explanation:

Step 1: Data given

Mass of iron = 60.0 grams

Initial temperature = 250 °C

Mass of gold = 60.0 grams

Initial temperature of gold = 45.0 °C

The specific heat capacity of iron = 0.449 J/g•°C

The specific heat capacity of gold = 0.128 J/g•°C.

Step 2: Calculate the final temperature at the equilibrium

Heat lost = Heat gained

Qlost = -Qgained

Qiron = -Qgold

Q=m*c*ΔT

m(iron) * c(iron) *ΔT(iron) = -m(gold) * c(gold) *ΔT(gold)

⇒with m(iron) = the mass of iron = 60.0 grams

⇒with c(iron) = the specific heat of iron = 0.449 J/g°C

⇒with ΔT(iron)= the change of temperature of iron = T2 - T1 = T2 - 250.0°C

⇒with m(gold) = the mass of gold= 60.0 grams

⇒with c(gold) = the specific heat of gold = 0.128 J/g°C

⇒with ΔT(gold) = the change of temperature of gold = T2 - 45.0 °C

60.0 *0.449 * (T2 - 250.0) = -60.0 * 0.128 * (T2 - 45.0 )

26.94 * (T2 - 250.0) = -7.68 * (T2 - 45.0)

26.94T2 - 6735 = -7.68T2 + 345.6

34.62T2 = 7080.6

T2 = 204.5 °C

The final temperature at the equilibrium is 204.6 °C

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The residual NaOH reacts with H₂SO₄ as follows:

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The equivalent-gram of H₂SO₄ are:

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As the complete residual solution is 100mL but the neutralization was made only with 10mL, the mEq you need to neutralize the residual NaOH is:

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The mEq of NaOH you add in the first are:

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That means the NaOH that reacts = moles of NH₄Cl is:

23.75mEq - 16.96mEq = 6.79mEq = 6.79mmoles NH₄Cl =

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therefore:

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