An ice tray contains 490 g of liquid water at 0°C. Calculate the change in entropy of the water as it freezes slowly and completely at 0°C.
i tired doing -(.49kg)(2.26x106)/273.15K) but im not getting the right answr help pls!
I don't know what the 2.26E6 is. Shouldn't you be using the heat of fusion of water? 333.55kJ/kg ?
To calculate the change in entropy during the freezing process, you need to consider the change in heat (energy) transferred and the temperature at which it happens. However, your equation seems to be incorrect.
The entropy change during the phase transition (freezing) of water can be calculated using the formula:
ΔS = mL/T
Where:
ΔS = change in entropy (in J/K)
m = mass of the substance (in kg)
L = latent heat of fusion (for water, it is approximately 2.26 x 10^6 J/kg)
T = temperature (in K)
In this case, the mass of water is given as 490 g, so we need to convert it to kilograms by dividing it by 1000:
m = 490 g / 1000 = 0.49 kg
The temperature is given as 0°C, which needs to be converted to Kelvin by adding 273.15:
T = 0°C + 273.15 = 273.15 K
Now, we can substitute these values into the formula:
ΔS = (0.49 kg) * (2.26 x 10^6 J/kg) / 273.15 K
Calculating this will give you the change in entropy during the freezing process.