Final Answer:
The final temperature of the gold and the water is 29.4°C.
Step-by-step explanation:
When two objects with different temperatures are placed in contact, they will exchange heat until they reach a common temperature, known as the equilibrium temperature. The amount of heat transferred between the two objects can be calculated using the following equation:
Q = mcΔT
where:
Q is the heat transferred (J)
m is the mass of the object (g)
c is the specific heat capacity of the object (J/g°C)
ΔT is the change in temperature (°C)
We can use this equation to calculate the heat transferred from the gold to the water. The specific heat capacity of gold is 0.128 J/g°C, and the specific heat capacity of water is 4.184 J/g°C.
Q_gold = (255.0 g) × (0.128 J/g°C) × (105°C - T)
Q_water = (155.0 g) × (4.184 J/g°C) × (T - 22.0°C)
Since the system is isolated, the heat transferred from the gold must equal the heat transferred to the water:
Q_gold = Q_water
Substituting the equations for Q_gold and Q_water, we get:
(255.0 g) × (0.128 J/g°C) × (105°C - T) = (155.0 g) × (4.184 J/g°C) × (T - 22.0°C)
Solving for T, we get:
T = 29.4°C
Therefore, the final temperature of the gold and the water is 29.4°C.