Answer:
a) 17.81 J/K
b) 33.325 J/K
Explanation:
The expression to use here is the following:
ΔS = Q/T
Where:
Q: heat released or absorbed
T: Temperature in K
Now, in order to do this, we need to gather the data. We know that the temperature in part a) is above the freezing temperature of water, which is 0 ° C or 273 K. and the mass of the ice cube is 14.6 g.
a) Using the water heat of fusion (Cause it's melting), we can calculated the heat released using the following expression:
Q = m * Lf
Lf = 333,000 J/kg
Solving for Q first we have:
Q = (14.6 / 1000) * 333,000
Q = 4,861.8 J
Now, the entropy change is:
ΔS = 4,861.8 / 273
ΔS = 17.81 J/K
b) In this part, we follow the same procedure than in part a) but using the water heat of boiling (Lv = 2,256,000 J/kg), the temperature of boiling which is 100 °C (or 373 K) and the mass of 5.51 g (0.00551 kg)
Calculating the heat:
Q = 0.00551 * 2,256,000 = 12,430.56 J
Now the entropy change:
ΔS = 12,430.56 / 373
ΔS = 33.325 J/K
