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Answer :
Final answer:
The final temperature of the aluminum and water system will be 23.5 °C.
Explanation:
To calculate the final temperature of the aluminum and water system, we can use the principle of heat transfer. The heat lost by the aluminum is equal to the heat gained by the water.
The equation used to calculate heat transfer is:
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.
For the aluminum:
QAl = (mass of aluminum) x (specific heat capacity of aluminum) x (final temperature - initial temperature of aluminum)
For the water:
Qwater = (mass of water) x (specific heat capacity of water) x (final temperature of water - initial temperature of water)
Since the system is completely isolated, the total heat lost by the aluminum is equal to the total heat gained by the water:
QAl = -Qwater
We can now substitute the known values into the equation:
(mass of aluminum) x (specific heat capacity of aluminum) x (final temperature - initial temperature of aluminum) = (mass of water) x (specific heat capacity of water) x (final temperature of water - initial temperature of water)
Rearranging the equation to solve for the final temperature:
final temperature = [(mass of aluminum) x (specific heat capacity of aluminum) x (initial temperature of aluminum) + (mass of water) x (specific heat capacity of water) x (initial temperature of water)] / [(mass of aluminum) x (specific heat capacity of aluminum) + (mass of water) x (specific heat capacity of water)]
Substituting the given values:
final temperature = [(3.90 g) x (0.897 J/g °C) x (99.3 °C) + (10.0 g) x (4.18 J/g °C) x (22.6 °C)] / [(3.90 g) x (0.897 J/g °C) + (10.0 g) x (4.18 J/g °C)]
Simplifying the equation:
final temperature = 23.5 °C
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Answer:
To determine the final temperature of the aluminum and water system, we can use the principle of conservation of energy, specifically the equation for heat transfer:
q_aluminum + q_water = 0
The heat gained by the aluminum (q_aluminum) is equal to the heat lost by the water (q_water) since there is no heat transfer to the surroundings.
The equation for heat transfer is:
q = m * c * ΔT
where q is the heat transfer, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
Given:
Mass of aluminum (m_aluminum) = 3.90 g
Initial temperature of aluminum (T_aluminum_initial) = 99.3 °C
Specific heat capacity of aluminum (c_aluminum) = 0.897 J/g°C
Volume of water (V_water) = 10.0 mL = 10.0 g (since the density of water is approximately 1 g/mL)
Initial temperature of water (T_water_initial) = 22.6 °C
Specific heat capacity of water (c_water) = 4.18 J/g°C
Using the equation for heat transfer, we can write:
(m_aluminum * c_aluminum * ΔT_aluminum) + (m_water * c_water * ΔT_water) = 0
Since the final temperature of the aluminum and water system will be the same, we can let ΔT_aluminum = ΔT_water = ΔT.
Substituting the values into the equation, we have:
(3.90 g * 0.897 J/g°C * ΔT) + (10.0 g * 4.18 J/g°C * ΔT) = 0
Simplifying the equation, we get:
3.49 ΔT + 41.8 ΔT = 0
45.29 ΔT = 0
Since the sum of the terms on the left side of the equation is zero, the change in temperature (ΔT) is zero. This means that the final temperature of the aluminum and water system will be the same as their initial temperatures.
Therefore, the final temperature of the aluminum and water will be:
T_final = T_aluminum_initial = 99.3 °C
The correct answer is:
b) Equal to 99.3 °C
