How much heat energy is released if 24.3 g of gold is cooled from 2580.0°C to room temperature (20.0°C)?

-8024.832 J

Explanation:

The amount of heat energy lost or released, Q, can be calculated thus;

Q = m × c × ∆T

Where;

Q = amount of heat lost (J)

m = mass of substance (g)

c = specific heat capacity (J/g°C)

∆T = change in temperature (°C)

According to this question,

Q = ?

m = 24.3 g

c of Gold = 0.129 J/g°C

∆T = 20°C - 2580.0°C = -2560°C

Using Q = m × c × ∆T

Q = 24.3 × 0.129 × -2560°C

Q = -8024.832 J

This means that 8024.832 J of energy is lost or released.

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frequency, z, and the collision density, z, in carbon dioxide (co2), r 180 pm at 30 °c and 120 kpa