The melting point of aluminum is 660°C, its latent heat of fusion is 4.00 × 105 J/kg, and its specific heat is 900 J/kg • K. If 300 kJ of heat are added to 442 g of aluminum at 100°C, what is the final state of the system? That is, how much is liquid, how much is solid, and what is its temperature?
(Please write answer in essay format)
Sum of heats gained is zero. Let M be the mass that is still solid.
heat added+heatmelting+heattoinceasetemp=0
300E3+(.442*900(660-100))+(.442-M)4E5=0
all are in that equation as Joules, Kg.
so solve the equation for M, the mass which is still solid at the melting point.
To determine the final state of the system, we need to consider the heat transfer and phase changes that occur when 300 kJ of heat are added to 442 g of aluminum at 100°C.
First, let's calculate the initial energy content of the aluminum using its specific heat. The equation for heat transfer is given by:
Q = mcΔT
where Q is the heat transferred, m is the mass, c is the specific heat, and ΔT is the change in temperature.
In this case, we have:
Q = 300 kJ = 300,000 J
m = 442 g = 0.442 kg
c = 900 J/kg • K
ΔT = final temperature - initial temperature = Tf - 100°C
Rearranging the equation, we can solve for ΔT:
ΔT = Q / (mc)
ΔT = 300,000 J / (0.442 kg * 900 J/kg • K) ≈ 746.5 K
Next, we need to determine how much heat is required to raise the aluminum from its initial temperature of 100°C to its melting point of 660°C. Since we know the specific heat and mass of the aluminum, we can apply the same equation as before:
Q = mcΔT
ΔT = Tf - 100°C = 660°C - 100°C = 560 K
Q = mcΔT = (0.442 kg) * (900 J/kg • K) * (560 K) ≈ 221,832 J
Now, let's consider the heat required for the phase change from solid to liquid. The equation for the heat transfer during a phase change is given by:
Q = mL
where Q is the heat transferred, m is the mass, and L is the latent heat of fusion.
In this case, Q = mL = (0.442 kg) * (4.00 × 105 J/kg) = 176,080 J
So, the total heat required for the aluminum to reach its melting point and completely melt is approximately 221,832 J + 176,080 J = 397,912 J.
Since this is larger than the initial heat of 300,000 J, we can conclude that the aluminum will completely melt. Additionally, the remaining heat will raise the temperature of the liquid aluminum as there is no further phase change after melting.
To determine the final temperature, we can rearrange the equation for heat transfer using specific heat:
Q = mcΔT
ΔT = Q / (mc) = (300,000 J - 397,912 J) / (0.442 kg * 900 J/kg • K) ≈ -109.57 K
The negative sign indicates a decrease in temperature. Subtracting this temperature change from the initial temperature of 100°C, we obtain:
Tf = 100°C - 109.57 K ≈ -9.57°C
Therefore, the final state of the system is completely melted aluminum with a temperature of approximately -9.57°C.
To determine the final state of the system, we need to consider the amount of heat added and the properties of aluminum.
1. Calculate the heat required to raise the temperature of the aluminum from 100°C to its melting point of 660°C:
Heat = mass × specific heat × change in temperature
Heat = 0.442 kg × 900 J/kg • K × (660°C - 100°C)
Heat = 244,716 kJ (or 2.44716 × 10^5 J)
2. Compare the amount of heat required to melt the aluminum with the heat added:
Heat required to melt the aluminum = mass × latent heat of fusion
Heat required to melt the aluminum = 0.442 kg × 4.00 × 10^5 J/kg
Heat required to melt the aluminum = 1.768 × 10^5 J
Since the heat added (300 kJ or 3.00 × 10^5 J) is greater than the heat required to melt the aluminum, we can conclude that all of the aluminum will melt.
3. Calculate the remaining heat after the aluminum has melted:
Remaining heat = Heat added - Heat required to melt the aluminum
Remaining heat = 3.00 × 10^5 J - 1.768 × 10^5 J
Remaining heat = 1.232 × 10^5 J (or 123,200 J)
4. Calculate the change in temperature due to the remaining heat after melting:
Change in temperature = Remaining heat / (mass × specific heat)
Change in temperature = 1.232 × 10^5 J / (0.442 kg × 900 J/kg • K)
Change in temperature = 307.8 K (or 34.8°C)
The final temperature of the system, after all the aluminum has melted, is 100°C + 34.8°C = 134.8°C.
In conclusion, when 300 kJ of heat are added to 442 g of aluminum at 100°C, all of the aluminum will melt. The final temperature will be 134.8°C, and the aluminum will be completely in the liquid state.