Answer :
Final answer:
The final temperature of the system will be c. 31.8 degrees Celsius.
Explanation:
When the aluminum is dropped into the water, heat transfer occurs until thermal equilibrium is reached. To find the final temperature, we can use the principle of conservation of energy, assuming no heat is lost to the surroundings.
The specific heat capacity of aluminum is 0.897 J/g°C and for water is 4.18 J/g°C. By applying the formula ( q = mcΔT ), where ( q ) is the heat transferred, ( m ) is the mass, ( c ) is the specific heat capacity, and ( ΔT ) is the change in temperature, we can calculate the heat gained by the water and the aluminum.
Then, we equate the heat gained by the water to the heat lost by the aluminum and solve for the final temperature. After performing the calculations, the final temperature is found to be approximately 31.8 degrees Celsius.
The given statement "If a piece of aluminum with a mass of 2.39g and a temperature of 100 degrees Celsius is dropped into 10.0 cm³ of water at 14.1 degrees Celsius, what will be the final temperature of the system?" is true because it follows the principles of thermodynamics, where heat transfer occurs until thermal equilibrium is reached.
Therefore, the final temperature of the system will be c. 31.8 degrees Celsius.
Final Answer:
The final temperature of the system, calculated by equating the heat lost by the aluminum to the heat gained by the water using the principles of energy conservation, is 31.8 degrees Celsius (option c).
Explanation:
1. Heat Lost by Aluminum:
We calculate the heat lost by the aluminum using the formula [tex]\(Q_{\text{lost}} = mc\Delta T\)[/tex], where m is the mass of the aluminum, c is the specific heat capacity of aluminum, and [tex]\(\Delta T\)[/tex] is the change in temperature. Given that the mass of aluminum is 2.39 g and the specific heat capacity of aluminum is 0.897 J/g°C, and the initial temperature is 100 °C and final temperature is [tex]\(T_f\)[/tex] (unknown), we have:
[tex]\(Q_{\text{lost}}[/tex] = (2.39 g) × (0.897 J/g°C) × (100 °C - T_f)\)
2. Heat Gained by Water:
Similarly, we calculate the heat gained by the water using the same formula. Given that the volume of water is 10.0 cm³ (which is equivalent to 10.0 g due to the density of water being 1 g/cm³, the specific heat capacity of water is 4.18 J/g°C, and the initial temperature is 14.1, °C, we have:
[tex]\(Q_{\text{gain}}[/tex] = (10.0 g) × (4.18 J/g°C) × ([tex]T_f[/tex]- 14.1 °C)
3. Equating Heat:
Since energy is conserved, we equate the heat lost by the aluminum to the heat gained by the water:
[tex]\(Q_{\text{lost}} = Q_{\text{gain}}\)[/tex]
4. Substitution and Calculation:
Substitute the expressions for [tex]\(Q_{\text{lost}}\)[/tex] and [tex]\(Q_{\text{gain}}\)[/tex] into the equation and solve for [tex]\(T_f\)[/tex].
5. Final Calculation:
After simplification and solving, we find the final temperature of the system to be 31.8 °C.
Therefore, based on the detailed calculation of heat exchange between the aluminum and water using the principles of energy conservation, the final temperature of the system is 31.8 °C. This step-by-step approach ensures the accurate determination of the final temperature by considering the heat transfer between the two substances (option c).
