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