100 grams of molten lead (600oC) is used to make musket balls. If the lead shot is allowed to cool to room temperature (21oC), what is the change in entropy (in J/K) of the lead? (For the specific heat of molten and solid lead use 1.29 J/g oC; the latent heat of fusion and the melting point of lead are 2.45 x 104 J/kg and 327oC.)
Check your given specific heat of lead. Yours appears to be off by a factor of about ten. I get 0.127 J/g*K for the solid and 0.120 J/g*K for the liquid, from various web sites.
To find the change in entropy, we need to consider the phase change from molten to solid lead and the change in temperature.
1. First, let's calculate the heat transferred during the phase change from molten to solid lead using the latent heat of fusion.
The latent heat of fusion is given as 2.45 x 10^4 J/kg. To find the heat transferred, we need to convert the mass of lead from grams to kilograms. So, 100 grams = 0.1 kg.
Heat transferred during the phase change = latent heat of fusion * mass of lead
= (2.45 x 10^4 J/kg) * 0.1 kg
= 2450 J
2. Next, let's calculate the heat transferred due to the change in temperature from 600oC to 21oC.
The specific heat capacity of molten and solid lead is given as 1.29 J/g°C.
Heat transferred due to the change in temperature = specific heat capacity * mass * change in temperature
For the change in temperature, we need to convert it from oC to Kelvin. The change in temperature is (21oC - 600oC) = -579oC. Converting this to Kelvin, we have -579 K.
Heat transferred due to the change in temperature = (1.29 J/g°C) * 100 g * (-579 K)
= -74991 J
Note: The negative sign indicates the heat loss.
3. Now, let's calculate the total change in entropy. Entropy change equals the total heat transferred divided by the temperature.
Entropy change = (heat transferred during phase change + heat transferred due to change in temperature) / temperature
We need to convert the temperatures from Celsius to Kelvin.
Initial temperature (molten lead) = 600oC + 273.15 K = 873.15 K
Final temperature (room temperature) = 21oC + 273.15 K = 294.15 K
Entropy change = (2450 J + (-74991 J)) / (294.15 K - 873.15 K)
= -72541 J / -579 K
≈ 125.40 J/K
Therefore, the change in entropy of the lead is approximately 125.40 J/K.
To calculate the change in entropy of the lead, we need to consider two aspects: the change in temperature and the change in phase.
First, let's calculate the change in temperature. We need to know the specific heat capacity of lead to determine the amount of heat transferred.
The specific heat capacity of molten lead is given as 1.29 J/g oC. So, for 100 grams of lead, the amount of heat transferred during the cooling process can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
ΔT = 600oC - 21oC = 579oC
Q = 100 g * 1.29 J/g oC * 579oC = 74691 J
Next, let's calculate the change in phase. Lead goes from a liquid phase to a solid phase, so we need to consider the latent heat of fusion.
The latent heat of fusion for lead is given as 2.45 x 10^4 J/kg. To calculate the amount of heat transferred during the phase change, we need to convert the mass of lead from grams to kilograms.
Mass of lead in kg = 100 g / 1000 = 0.1 kg
Q = m * L
where Q is the heat transferred, m is the mass of the substance, and L is the latent heat of fusion.
Q = 0.1 kg * 2.45 x 10^4 J/kg = 2450 J
Now, let's calculate the total change in entropy.
ΔS = Q / T
where ΔS is the change in entropy, Q is the total heat transferred, and T is the temperature in Kelvin.
The final temperature after cooling to room temperature is 21oC, which is 21 + 273 = 294 K.
ΔS = (74691 J + 2450 J) / 294 K = 253 J/K
Therefore, the change in entropy of the lead is 253 J/K.
For each incremental heat loss dQ, there is an entropy loss of the lead that is equal to dQ/T, where T must be in Kelvin. Compute the total entropy loss in three steps:
(1) liquid lead cooling from 873 to 600 C
(2) molten lead turning to solid at a contant temperature of 600 K while the heat of fusion is transferred away, and
(3) solid lead cooling from 600 K to 294 K.
You are supposed to use the same specific heats for the liquid and solid lead in this case; this is an approximation. Call the specific heat of both C.
Entropy loss is:
(1) Integral of dQ/T from T = 873 to 600 = M C ln(873/600), where M is the mass and C is the specific heat,PLUS
(2) Integral of dQ/T as the lead freezes, M*Qf/600, where Qf is the latent heat of fusion, PLUS
(3) Integral of dQ/T from T=600 to T = 294 K, = M C ln(600/400)
Now just add up the numbers. Use numeric values for M, Qf and C.