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 that occurs when the water and ice mixture reaches equilibrium, we need to calculate the change in entropy for both the heating of water and the melting of ice separately.
1. Change in entropy for heating water:
- Given: mass of water (m) = 6.00 kg, initial temperature (Ti) = 80.4 °C
- We need to convert the initial temperature to Kelvin by adding 273.15: Ti = 80.4 + 273.15 = 353.55 K
- Final temperature (Tf) will be the equilibrium temperature of the mixture.
- We can use the specific heat capacity of water, which is approximately 4186 J/kg·K.
- Using the provided expression, ÄS = mc ln(Tf/Ti):
ΔS_water = 6.00 kg × 4186 J/kg·K × ln(Tf/353.55 K)
2. Change in entropy for melting ice:
- Given: mass of ice (m) = 3.00 kg
- The change in entropy for melting ice, ΔS_ice, can be calculated using the formula:
ΔS_ice = m × ΔH_fusion / T
where ΔH_fusion is the latent heat of fusion for water, which is approximately 334000 J/kg, and T is the melting point of ice, 0.0 °C.
- We need to convert the temperature to Kelvin: T = 0.0 + 273.15 = 273.15 K
- Using the given values:
ΔS_ice = 3.00 kg × 334000 J/kg / 273.15 K
3. Total change in entropy:
- The total change in entropy, ΔS_total, will be the sum of the change in entropy for heating water and the change in entropy for melting ice:
ΔS_total = ΔS_water + ΔS_ice
Now, you can substitute the values you know into the equations to calculate the change in entropy for each part and the total change in entropy.