A hot lump of 25.6 g of aluminum at an initial temperature of 57.4 °C is placed in 50.0 mL of H2O initially at 25.0 °C and allowed to reach thermal equilibrium. What is the final temperature of the aluminum and water given that the specific heat of aluminum is 0.903 J/(g·°C)? Assume no heat is lost to surroundings.
Answer in °C
Review 'Method of Mixtures' type problems.
To find the final temperature of the aluminum and water when they reach thermal equilibrium, we can use the principle of conservation of energy, specifically the principle of heat transfer. The heat gained by the aluminum will be equal to the heat lost by the water.
The heat gained by the aluminum can be calculated using the formula:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
Where:
Q_aluminum = heat gained by aluminum
m_aluminum = mass of aluminum
c_aluminum = specific heat capacity of aluminum
ΔT_aluminum = change in temperature of aluminum
And the heat lost by the water can be calculated using the formula:
Q_water = m_water * c_water * ΔT_water
Where:
Q_water = heat lost by water
m_water = mass of water
c_water = specific heat capacity of water
ΔT_water = change in temperature of water
Since the heat gained and heat lost are the same when the aluminum and water reach thermal equilibrium, we can equate the two equations:
Q_aluminum = Q_water
m_aluminum * c_aluminum * ΔT_aluminum = m_water * c_water * ΔT_water
We are given the following values:
m_aluminum = 25.6 g
c_aluminum = 0.903 J/(g·°C)
ΔT_aluminum = final temperature of aluminum - initial temperature of aluminum
m_water = 50.0 mL (which is equivalent to 50.0 g, assuming the density of water is 1 g/mL)
c_water = 4.18 J/(g·°C) (specific heat capacity of water)
ΔT_water = final temperature of water - initial temperature of water
Since the final temperatures of aluminum and water are the same at thermal equilibrium, we can denote it as T_f.
Now, let's solve the equation:
m_aluminum * c_aluminum * ΔT_aluminum = m_water * c_water * ΔT_water
25.6 g * 0.903 J/(g·°C) * (T_f - 57.4 °C) = 50.0 g * 4.18 J/(g·°C) * (T_f - 25.0 °C)
Simplifying the equation:
23.1744 (T_f - 57.4) = 209 J/(g·°C) (T_f - 25.0)
23.1744 T_f - 2438.57 = 209 T_f - 5225
Rearranging the equation:
23.1744 T_f - 209 T_f = -5225 + 2438.57
-185.8256 T_f = -2786.43
Solving for T_f:
T_f = -2786.43 / -185.8256
T_f ≈ 15.0 °C
Therefore, the final temperature of the aluminum and water when they reach thermal equilibrium is approximately 15.0 °C.