After 6.00 kg of water at 80.4 oC is mixed in a perfect thermos with 3.00 kg of ice at 0.0 oC, the mixture is allowed to reach equilibrium. When heat is added to or removed from a solid or liquid of mass m and specific heat capacity c, the change in entropy can be shown to be ÄS = mc ln(Tf/Ti), where Ti and Tf are the initial and final Kelvin temperatures. Using this expression and the change in entropy for melting, find the change in entropy that occurs.
To find the change in entropy, we need to consider the heat exchange during the process. First, let's calculate the heat exchanged during the heating of water:
q1 = mcΔT
where:
m = mass of water = 6.00 kg
c = specific heat capacity of water = 4.18 kJ/kg·°C (at 80.4 °C)
ΔT = change in temperature = (final temperature of water - initial temperature of water) = (0.0 °C - 80.4 °C)
Note: We need to convert the temperature to Kelvin, as the formula requires it.
ΔT = (0.0 °C - 80.4 °C) = -80.4 °C = -80.4 K
Now, we plug in the values into the equation:
q1 = mcΔT = (6.00 kg)(4.18 kJ/kg·°C)(-80.4 K)
Next, let's calculate the heat exchanged during the phase change from ice to water:
q2 = mLf
where:
m = mass of ice = 3.00 kg
Lf = latent heat of fusion = 334 kJ/kg
Now, let's calculate the final temperature of the mixture. At equilibrium, the final temperature will be the same for both water and ice:
ΔS = mc ln(Tf/Ti)
Since ΔS = 0 (because the process is at equilibrium), we can rearrange the equation:
mc ln(Tf/Ti) = 0
Dividing both sides by mc:
ln(Tf/Ti) = 0
Taking the exponential of both sides:
Tf/Ti = e^0 = 1
So, Tf = Ti
Therefore, the final temperature of the mixture is 80.4 °C.
To find the change in entropy, we need to calculate the entropy change in each step and sum them up:
ΔS = ΔS1 + ΔS2
For the heating of water, we use the equation given:
ΔS1 = mc ln(Tf/Ti) = (6.00 kg)(4.18 kJ/kg·°C) ln(80.4 K/80.4 K)
For the phase change from ice to water, we use the formula:
ΔS2 = q2/Ti = (3.00 kg)(334 kJ/kg)/80.4 K
Now, you can calculate the value of ΔS1 and ΔS2, and then sum them up to obtain the total entropy change, ΔS.
To find the change in entropy that occurs when the water and ice mixture reaches equilibrium, we can use the given expression for the change in entropy and the change in entropy for melting.
Step 1: Calculate the change in entropy for the water.
Given:
- Mass of water (m) = 6.00 kg
- Specific heat capacity of water (c) = the specific heat capacity of water is approximately 4.18 J/g°C, or 4.18 kJ/kg°C (1 kg = 1000 g)
- Initial temperature of water (Ti) = 80.4 °C (convert to Kelvin by adding 273.15)
Convert the mass and specific heat capacity to joules (J) using the formula:
Mass (kg) * Specific heat capacity (kJ/kg°C) * 1000 = Specific heat capacity (J/°C)
6.00 kg * 4.18 kJ/kg°C * 1000 = 25,080 J/°C
Now, apply the change in entropy formula:
ΔS1 = m * c * ln(Tf / Ti)
where Tf is the final temperature in Kelvin. Since the water and ice mixture reaches equilibrium, Ti is the initial temperature of the water (80.4 °C converted to Kelvin), and Tf is the final temperature of the mixture.
ΔS1 = 6.0 kg * 25,080 J/°C * ln(Tf / (80.4 °C + 273.15 K))
Step 2: Calculate the change in entropy for the ice melting.
Given:
- Mass of ice (m) = 3.00 kg
- The change in entropy for melting is given as ΔSm = m * L, where L is the latent heat of fusion for ice.
The latent heat of fusion for ice is approximately 334 kJ/kg.
ΔSm = 3.00 kg * 334 kJ/kg = 1002 kJ
Convert the change in entropy for melting from kilojoules (kJ) to joules (J):
1002 kJ * 1000 J/1 kJ = 1,002,000 J
Step 3: Calculate the total change in entropy.
The change in entropy that occurs is the sum of the change in entropy for the water and the change in entropy for melting the ice:
ΔS_total = ΔS1 + ΔSm
Substitute the values we calculated earlier:
ΔS_total = (6.0 kg * 25,080 J/°C * ln(Tf / (80.4 °C + 273.15 K))) + 1,002,000 J
This equation gives you the change in entropy that occurs when the water and ice mixture reaches equilibrium.