A 2.00-kg block of ice is at STP (0¢XC, 1 atm) while it melts completely to water. What is its change in entropy? (For ice, Lf = 3.34 „e 105 J/kg)
The entropy change, which is an increase in this case, equals the heat of fusion of water divided by the absolute temperature, 273 K.
I do not understand your
¢X and „e symbols. You should precede exponents with a ^
The c with the line through it is a degree symbol, and the e is a ^ symble. I am sorry the computer put it in that way.
You have to multiply by the number of kg also. (heat of fusion)/T gives you the entropy per unit mass
3.34*10^5 J/kg * 2.00 kg /273 K = 2450 J/K, to three significant figures
Thank You
Why did we divide by absolute temp instead of 0C? @drwls
To find the change in entropy of the ice block as it melts completely to water, you can use the formula:
ΔS = Q / T
where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.
In this case, we need to consider the heat required to melt the ice, which is given by the formula:
Q = m * Lf
where Q is the heat, m is the mass of the ice block, and Lf is the heat of fusion.
Given that the mass of the ice block is 2.00 kg and the heat of fusion of ice is Lf = 3.34 × 10^5 J/kg, we can calculate the heat required:
Q = 2.00 kg * 3.34 × 10^5 J/kg
Q = 6.68 × 10^5 J
Next, we need to determine the initial and final temperatures. Since the ice is at 0°C (273.15 K), and it is undergoing a phase change to water, we can assume that the final temperature is also 0°C (273.15 K).
Now we can substitute the values into the formula for entropy change:
ΔS = Q / T
ΔS = 6.68 × 10^5 J / 273.15 K
Calculating this gives us the change in entropy for the ice block as it melts completely to water.