Answer:
The specific heat capacity of aluminum can be calculated using the equation q = m * c * ΔT. The heat transferred from boiling water to the aluminum bar is equal to the heat transferred from the aluminum bar to the room temperature water. By substituting the given values, the specific heat capacity of aluminum is calculated to be 63.1 J/g°C.
Step-by-step explanation:
To solve this problem, we can use the equation:
q = m * c * ΔT
where q is the heat transferred, m is the mass of the material, c is the specific heat capacity, and ΔT is the change in temperature.
We can first use this equation to calculate the heat transferred when the aluminum bar is heated in boiling water:
q1 = m1 * cAl * ΔT1
where m1 is the mass of the aluminum bar, cAl is the specific heat capacity of aluminum, and ΔT1 is the change in temperature of the aluminum bar as it reaches thermal equilibrium with the boiling water.
We know that the initial temperature of the aluminum bar is 100.0°C and the final temperature is 100.0°C (because it reached equilibrium with the boiling water). Therefore:
ΔT1 = 0°C
We can also calculate the heat transferred from the boiling water to the aluminum bar using the equation:
q1 = -qwater
where qwater is the heat transferred from the boiling water to the aluminum bar. We can calculate qwater using the equation:
qwater = mwater * cwater * ΔT1
where mwater is the mass of the boiling water, cwater is the specific heat capacity of water, and ΔT1 is the change in temperature of the boiling water as it reaches thermal equilibrium with the aluminum bar. We know that the initial temperature of the boiling water is 100.0°C and the final temperature is 100.0°C (because it reached equilibrium with the aluminum bar). Therefore:
ΔT1 = 0°C
Substituting in the given values, we get:
qwater = (236.0 g) * (4.184 J/g°C) * (0°C) = 0 J
Therefore:
q1 = -qwater = 0 J
This means that the heat transferred to the aluminum bar from the boiling water is equal to the heat transferred from the aluminum bar to the room temperature water.
We can use the equation again to calculate the heat transferred from the aluminum bar to the room temperature water:
q2 = m2 * cwater * ΔT2
where m2 is the mass of the room temperature water, cwater is the specific heat capacity of water, and ΔT2 is the change in temperature of the room temperature water as it reaches thermal equilibrium with the aluminum bar.
We know that the initial temperature of the room temperature water is 21.0°C and the final temperature is 23.0°C. Therefore:
ΔT2 = 23.0°C - 21.0°C = 2.0°C
Substituting in the given values, we get:
q2 = (236.0 g) * (4.184 J/g°C) * (2.0°C) = 1970.048 J
Therefore:
q1 = -q2
0 J = -1970.048 J
This means that the specific heat capacity of aluminum can be calculated as:
cAl = q2 / (m1 * ΔT1)
Substituting in the given values, we get:
cAl = (1970.048 J) / (31.2 g * 0°C) = 63.1 J/g°C
Therefore, the specific heat capacity of aluminum is 63.1 J/g°C.