A 25.0 g aluminum block is warmed to 65.2 C and plunged into an insulated beaker containing 55.3g of water initially at 22.0C. the aluminum and water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of the water and aluminum?
25.8 degrees Celsius
6742.8
Well, let's see here. We have an aluminum block and water competing to be the hottest contestants in an insulated beaker. It's like a sauna, but with elements instead of people.
First, we need to find out the heat gained by the water. We can use the equation Q = mcΔT, where Q is the heat gained, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.
For the water, we have:
Q_water = (55.3g) * (4.18 J/g°C) * (T_final - 22.0°C)
Now, for the aluminum block, we need to find the heat lost. Let's assume the final temperature is T_final for both substances. Using the same equation, we have:
Q_aluminum = -(25.0g) * (0.897 J/g°C) * (T_final - 65.2°C)
Since no heat is lost or gained from the surroundings, the heat gained by the water is equal to the heat lost by the aluminum block. Therefore:
Q_water = -Q_aluminum
Now we can set up the equation:
(55.3g) * (4.18 J/g°C) * (T_final - 22.0°C) = (25.0g) * (0.897 J/g°C) * (T_final - 65.2°C)
Now, let's solve this equation and find the final temperature. It's time to turn up the heat, or in this case, the math:
(55.3 * 4.18 * T_final) - (55.3 * 4.18 * 22.0) = (25.0 * 0.897 * T_final) - (25.0 * 0.897 * 65.2)
After doing the calculations, the final temperature of the water and aluminum should be around 24.5°C. So, it's decided - the water and aluminum have reached a compromising temperature in their cozy little insulated beaker.
To find the final temperature of the water and aluminum, we need to use the principle of conservation of energy and apply the formula for heat transfer.
The formula for heat transfer is given by:
Q = mcΔT
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.
First, let's calculate the heat transferred from the aluminum to the water. We'll assume that no heat is lost during the transfer.
The specific heat capacity of aluminum is 0.897 J/g°C, and the specific heat capacity of water is 4.184 J/g°C.
1. Calculate the heat transferred from the aluminum block:
Q_aluminum = m_aluminum * c_aluminum * ΔT_aluminum
m_aluminum = 25.0 g (mass of aluminum)
c_aluminum = 0.897 J/g°C (specific heat capacity of aluminum)
ΔT_aluminum = Tf_aluminum - Ti_aluminum
Tf_aluminum is the final temperature of the aluminum (what we want to find)
Ti_aluminum is the initial temperature of the aluminum, which is 65.2°C
2. Calculate the heat transferred to the water:
Q_water = m_water * c_water * ΔT_water
m_water = 55.3 g (mass of water)
c_water = 4.184 J/g°C (specific heat capacity of water)
ΔT_water = Tf_water - Ti_water
Tf_water is the final temperature of the water (what we want to find)
Ti_water is the initial temperature of the water, which is 22.0°C
Now, since the heat transferred from the aluminum to the water is equal, we can set up an equation:
Q_aluminum = Q_water
m_aluminum * c_aluminum * ΔT_aluminum = m_water * c_water * ΔT_water
Substituting the given values:
25.0 g * 0.897 J/g°C * (Tf_aluminum - 65.2°C) = 55.3 g * 4.184 J/g°C * (Tf_water - 22.0°C)
Simplifying the equation:
22.425 (Tf_aluminum - 65.2) = 231.118 (Tf_water - 22.0)
Now, we can solve this equation to find the final temperature of the water and aluminum.
the sum of heats gained is zero (one will give off heat, or negative heat gained)
massal*specificheatAl*(Tf-65.2)+masswater*c*(Tf-22)=0
solve for Tf