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.

Question

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.

Answer 1

To find the change in entropy that occurs when the water and ice mixture reaches equilibrium, we need to calculate the change in entropy due to heating the water and melting the ice separately.

Let's start with the water:
1. First, convert the initial temperature from Celsius to Kelvin by adding 273.15: Ti_water = 80.4 °C + 273.15 = 353.55 K
2. Next, calculate the final temperature (which will be the same for the water and ice mixture) by assuming it reaches equilibrium: Tf_water = Tf_ice = Tf
3. Now we can use the formula for the change in entropy due to heating:
ΔS_water = (mass_water) * (specific heat capacity_water) * ln(Tf_water / Ti_water)
= (6.00 kg) * (c_water) * ln(Tf_water / 353.55 K)

Next, let's consider the ice:
1. To calculate the change in entropy due to melting, we use the formula:
ΔS_ice = (mass_ice) * (latent heat of fusion) / (absolute temperature)
= (3.00 kg) * (Lf) / (Tf_ice)

Note: Lf is the latent heat of fusion for ice, which is the amount of heat required to convert one kg of ice at 0.0 °C to water at 0.0 °C. Its value is 334,000 J/kg.

Since the final temperature is the same for both the water and ice mixture, we can simplify the expressions:
ΔS_water = (6.00 kg) * (c_water) * ln(Tf / 353.55 K)
ΔS_ice = (3.00 kg) * (334,000 J/kg) / (Tf)

Finally, to find the total change in entropy for the water and ice mixture, you can sum the individual changes:
ΔS_total = ΔS_water + ΔS_ice

Please note that the specific heat capacity (c_water) for water should be known and the units must be consistent throughout the calculations. Additionally, the final temperature (Tf) is still unknown, so you would need additional information or equations to solve for it.