A 51.1-g sample of aluminum at 95.0 degree C is dropped into 35.0 g of water at 40.0 degree C --What is the final temperature of the mixture?
To find the final temperature of the mixture, we can use the principle of heat transfer. The amount of heat lost by the aluminum and gained by the water will be equal when they reach thermal equilibrium.
The heat lost by aluminum (q1) can be calculated using the formula:
q1 = m1 * C1 * ΔT1
Where:
m1 = mass of aluminum (51.1 g)
C1 = specific heat capacity of aluminum (0.897 J/g°C)
ΔT1 = change in temperature of aluminum (final temperature - initial temperature)
The heat gained by water (q2) can be calculated using the formula:
q2 = m2 * C2 * ΔT2
Where:
m2 = mass of water (35.0 g)
C2 = specific heat capacity of water (4.18 J/g°C)
ΔT2 = change in temperature of water (final temperature - initial temperature)
Since the heat lost by the aluminum is equal to the heat gained by the water, we can set q1 equal to q2:
m1 * C1 * ΔT1 = m2 * C2 * ΔT2
Now, we can solve this equation for the final temperature (Tf):
(Tf - 95.0°C) = ((m2 * C2) / (m1 * C1)) * (95.0°C - 40.0°C)
Finally, we can rearrange the equation to solve for Tf:
Tf = ((m2 * C2 * ΔT1) / (m1 * C1)) + 95.0°C
Plugging in the given values:
Tf = ((35.0 g * 4.18 J/g°C * (40.0°C - Tf)) / (51.1 g * 0.897 J/g°C)) + 95.0°C
Solving this equation will give us the final temperature of the mixture.