If 400 gm of aluminum are put into 200 gm of water at 20 C in 110 gr. calorimeter of specific heat .093, and the temp.or stable equilibrium is 28.8 C, what is the initial temperature of the aluminum? specific heat of aluminum is .22..
To find the initial temperature of the aluminum, we can use the principle of conservation of energy. The energy lost by the aluminum as it cools down is equal to the energy gained by the water and the calorimeter.
First, let's find the energy lost by the aluminum. We can use the formula:
Q = m * c * ΔT
Where:
Q is the amount of heat transferred
m is the mass of the substance (aluminum)
c is the specific heat capacity of the substance (aluminum)
ΔT is the change in temperature
In this case, the mass of aluminum is 400 grams, the specific heat capacity of aluminum is 0.22, and the change in temperature is ΔT = T_initial - T_eqilibrium.
So the energy lost by the aluminum is:
Q_aluminum = 400g * 0.22 * ΔT
Next, let's find the energy gained by the water and the calorimeter. We can use the same formula:
Q = m * c * ΔT
In this case, the mass of water is 200 grams, the specific heat capacity of water is 4.18 (approximate value), and the change in temperature is ΔT = T_eqilibrium - T_initial.
So the energy gained by the water and the calorimeter is:
Q_water = (200g + 110g) * 4.18 * ΔT
Now, according to the principle of conservation of energy, the energy lost by the aluminum is equal to the energy gained by the water and the calorimeter:
Q_aluminum = Q_water
Substituting the respective equations for Q_aluminum and Q_water, we get:
400g * 0.22 * ΔT = (200g + 110g) * 4.18 * ΔT
Now, we can cancel out ΔT from both sides and solve for the initial temperature T_initial:
400g * 0.22 = (200g + 110g) * 4.18
88g = 310g * 4.18
Dividing both sides by 310g * 4.18, we get:
88g / (310g * 4.18) = T_initial
T_initial ≈ 0.063
Therefore, the initial temperature of the aluminum is approximately 0.063 degrees Celsius.