If a chalice made of pure gold weighing 1.24 kg and initially at 15 C is filled with 100.0 mL of water at 89 C, what is the final equilibrium temperature of the system? The heat capacities of water and gold are 75.29 and 25.42 J/mol/K, respectively.

heat gained by gold + heat lost by water = 0

[mass Au x specific heat Au x (Tfinal-Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial) = 0
Solve for Tfinal.
Note that you quote specific heat of Au and H2O in J/mol; therefore, the mass of Au and mass H2O must be in moles.

To find the final equilibrium temperature of the system, we can use the principle of heat transfer, which states that the heat lost by one substance is equal to the heat gained by the other substance when they reach thermal equilibrium.

First, let's calculate the heat lost by the water. We can use the following equation:

Qwater = m * Cp * ΔT

Where:
- Qwater is the heat lost by the water
- m is the mass of the water
- Cp is the specific heat capacity of water
- ΔT is the change in temperature of the water

The mass of the water can be determined by converting the volume to mass using the density of water, which is approximately 1 g/mL:

m = V * density
m = 100.0 mL * 1 g/mL
m = 100.0 g

ΔT is the difference between the initial temperature of the water (89 °C) and the final temperature of the system.

ΔT = (final temperature) - (initial temperature)
ΔT = (final temperature) - (89 °C)

Now, let's calculate the heat lost by the water. Given that the specific heat capacity of water is 75.29 J/mol/K, we need to convert the mass of water to moles:

n = mass / molar mass of water
n = 100.0 g / 18.015 g/mol
n ≈ 5.549 mol

Qwater = n * Cp * ΔT
Qwater = 5.549 mol * 75.29 J/mol/K * (final temperature - 89 °C)

Next, let's calculate the heat gained by the gold. The heat gained can be determined using the same equation, but with the heat capacity of gold (25.42 J/mol/K) and the mass of the gold (1.24 kg).

Qgold = m * Cp * ΔT
Qgold = 1.24 kg * 25.42 J/mol/K * (final temperature - 15 °C)

Since the total heat lost by the water is equal to the heat gained by the gold, we can set the two equations equal to each other:

Qwater = Qgold

5.549 mol * 75.29 J/mol/K * (final temperature - 89 °C) = 1.24 kg * 25.42 J/mol/K * (final temperature - 15 °C)

Now, we can solve for the final temperature of the system by isolating the variable:

5.549 * 75.29 * (final temperature - 89) = 1.24 * 25.42 * (final temperature - 15)

After simplifying and rearranging the equation, we get:

78.6571 * (final temperature - 89) = 31.5448 * (final temperature - 15)

78.6571 * final temperature - 7013.57 = 31.5448 * final temperature - 473.172

(78.6571 - 31.5448) * final temperature = 7013.57 - 473.172

47.1123 * final temperature = 6540.398

final temperature = 6540.398 / 47.1123
final temperature ≈ 138.86 °C

Therefore, the final equilibrium temperature of the system is approximately 138.86 °C.