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Answer :
Final answer:
The final temperature of the water and metal when a 16.8 g sample of iron is placed in 150.0 mL of water is 25.0 °C.
Explanation:
When two substances with different temperatures are brought into contact, they will eventually reach the same temperature through a process called thermal equilibrium. In this scenario, the iron sample is initially at a temperature of 98.2 °C and the water is at 25.0 °C. The final temperature will be the average of these two temperatures, which is 61.6 °C ((98.2+25.0)/2). However, this is not the final answer as the iron sample will transfer its heat to the water until they reach thermal equilibrium.
To calculate the final temperature, we can use the formula Q = m x Cs x ΔT, where Q represents the heat transferred, m is the mass, Cs is the specific heat, and ΔT is the change in temperature. The heat transferred by the iron sample can be calculated as Qiron = (16.8 g)(0.449 J/g°C)(98.2 °C - Tfinal). Similarly, the heat transferred by the water can be calculated as Qwater = (150.0 mL)(1.00 g/mL)(4.18 J/g°C)(Tfinal - 25.0 °C).
Since the heat lost by the iron is equal to the heat gained by the water (assuming no heat is lost to the surroundings), we can set Qiron equal to Qwater and solve for Tfinal. This gives us the equation (16.8 g)(0.449 J/g°C)(98.2 °C - Tfinal) = (150.0 mL)(1.00 g/mL)(4.18 J/g°C)(Tfinal - 25.0 °C). Solving for Tfinal gives us a final temperature of 25.0 °C, which is the correct answer (C) listed in the question.
In conclusion, the final temperature of the water and iron sample is 25.0 °C. This is because heat is transferred from the iron sample to the water until they reach thermal equilibrium, and the final temperature is the average of the initial temperatures. This can be calculated using the formula Q = m x Cs x ΔT and setting the heat lost by the iron equal to the heat gained by the water. Following these calculations, we can determine the correct answer (C) for the given question.
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