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.
q(Al) + q(HOH) = 0
mcT(Al) + mcT(HOH)= 0
[(25)(0.903)(T-57.4)] + [(25)(4.184)(T-25) = 0
Solve for T.
To find the final temperature of the aluminum and water mixture, we can use the principle of heat transfer:
q(aluminum) + q(water) = 0
The equation represents the conservation of energy principle, stating that the heat gained by the water must be equal to the heat lost by the aluminum.
The heat gained or lost by an object is given by the equation:
q = m * c * ΔT
where q is the heat gained or lost, m is the mass of the object, c is the specific heat capacity of the object, and ΔT is the change in temperature.
For the aluminum:
q(aluminum) = m(aluminum) * c(aluminum) * ΔT(aluminum)
For the water:
q(water) = m(water) * c(water) * ΔT(water)
In our case, we have:
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 g (since the density of water is 1 g/mL, and the volume is given as 50.0 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 aluminum and water reach thermal equilibrium, their final temperatures are the same. Let's call it Tf.
Now we can write the equation for heat transfer as:
m(aluminum) * c(aluminum) * ΔT(aluminum) + m(water) * c(water) * ΔT(water) = 0
Now let's plug in the given values:
25.6 g * 0.903 J/(g·°C) * (Tf - 57.4 °C) + 50.0 g * 4.18 J/(g·°C) * (Tf - 25.0 °C) = 0
Simplifying the equation:
23.136 g·°C * (Tf - 57.4 °C) + 209 J/°C * (Tf - 25.0 °C) = 0
Now we can solve for Tf:
23.136 g·°C * Tf - 23.136 g·°C * 57.4 °C + 209 J/°C * Tf - 209 J/°C * 25.0 °C = 0
(23.136 g·°C + 209 J/°C) * Tf = 23.136 g·°C * 57.4 °C + 209 J/°C * 25.0 °C
(23.136 g·°C + 209 J/°C) * Tf = 1329.9264 g·°C + 5225 J/°C
Tf = (1329.9264 g·°C + 5225 J/°C) / (23.136 g·°C + 209 J/°C)
Calculating Tf:
Tf = 65.64 °C
Therefore, the final temperature of the aluminum and water mixture is 65.64 °C.
To find the final temperature of the aluminum and water mixture, we can use the principle of conservation of energy. The heat lost by the aluminum equal to the heat gained by the water.
To calculate the heat lost by the aluminum, we can use the formula:
Q_aluminum = m × c × ΔT
Where:
Q_aluminum = heat lost by aluminum
m = mass of aluminum (in grams)
c = specific heat of aluminum (in J/(g·°C))
ΔT = change in temperature of aluminum (final temperature - initial temperature)
Given:
m = 25.6 g
c = 0.903 J/(g·°C)
ΔT = final temperature of aluminum - initial temperature of aluminum
To calculate the heat gained by the water, we can use the formula:
Q_water = m × c × ΔT
Where:
Q_water = heat gained by water
m = mass of water (in grams)
c = specific heat of water (assumed to be 4.18 J/(g·°C) for water)
ΔT = change in temperature of water (final temperature - initial temperature)
Given:
m = 50.0 mL (convert to grams)
c = 4.18 J/(g·°C)
ΔT = final temperature of water - initial temperature of water
Since the heat lost by the aluminum is equal to the heat gained by the water, we can set up the equation:
Q_aluminum = Q_water
m × c_aluminum × ΔT_aluminum = m × c_water × ΔT_water
Now let's plug in the given values and solve for the final temperature of aluminum and water.
1. Convert the mass of water from milliliters to grams:
m_water = 50.0 mL × 1 g/mL = 50.0 g
2. Solve for ΔT_aluminum:
ΔT_aluminum = final temperature of aluminum - initial temperature of aluminum
3. Solve for ΔT_water:
ΔT_water = final temperature of water - initial temperature of water
4. Set up the equation and solve for the final temperatures:
m_aluminum × c_aluminum × ΔT_aluminum = m_water × c_water × ΔT_water
25.6 g × 0.903 J/(g·°C) × ΔT_aluminum = 50.0 g × 4.18 J/(g·°C) × ΔT_water
25.6 × 0.903 × ΔT_aluminum = 50.0 × 4.18 × ΔT_water
Now you can rearrange the equation and solve for the final temperature of the aluminum and water.