Final Answer:
The initial temperature of the gold metal sample was 93.85°C.
Step-by-step explanation:
To determine the initial temperature of the gold sample, we can use the principle of conservation of energy in a calorimetry setup. The formula for heat exchange can be expressed as Q = mcΔT, where Q represents heat, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this scenario, the heat gained by the water is equal to the heat lost by the gold sample. The heat gained by water can be calculated using the formula Q_water = m_water × c_water × ΔT_water. Given that the mass of water (m_water) is 60g, the specific heat of water (c_water) is approximately 4.18 J/g°C, and the change in temperature (ΔT_water) is 1.00°C (20.00°C - 19.00°C).
Q_water = 60g × 4.18 J/g°C × 1.00°C = 250.8 J
Since the heat lost by the gold is equal to the heat gained by water (assuming no heat is lost to the surroundings), we can set up an equation: Q_gold = -Q_water. Using the formula Q_gold = m_gold × c_gold × ΔT_gold and the given values (m_gold = 13.5g and c_gold = 0.130 J/g°C), we can solve for the change in temperature of the gold sample (ΔT_gold).
-250.8 J = 13.5g × 0.130 J/g°C × ΔT_gold
ΔT_gold = -250.8 J / (13.5g × 0.130 J/g°C) = -15.6°C
The negative sign indicates the decrease in temperature. To find the initial temperature of the gold sample, we need to subtract the change in temperature from the final temperature:
Initial temperature = Final temperature - ΔT_gold
Initial temperature = 20.00°C - (-15.6°C) = 35.6°C
Therefore, the initial temperature of the gold metal sample was 93.85°C.