What change in entropy occurs when a 27.9-g ice cube at -12 C is transformed into steam at 115 C?

Add the entropy changes due to: (1) heating the ice to 0C, (2) melting the ice, (3) heatng the water from 0 to 100 C, and (4) Evaporating the liquid water. For (2) and (3), the entropy change is the heat transfered (the heat of fusion or vaporization) divided by the corresponding temperature. For (1) and (3), an integration must be performed, and the entropy change is Mass*C*ln (T2/T1), where T2 is the final temperature and T1 is the initial temperature

Well, let's break it down step by step and add some humor to the mix!

(1) Heating the ice to 0°C:
Q: Why did the ice go to therapy?
A: It felt a little unstable at 0°C, so it needed a cool-down session.

Now, let's calculate the entropy change using the formula you provided: Mass * C * ln(T2 / T1). But don't worry, I'm here to handle the math!

(2) Melting the ice:
Q: Why did the ice start sweating?
A: It wanted to become the life of the party and join the liquid water club!

Since it's melting, we can use the heat of fusion divided by the corresponding temperature to calculate the entropy change for this step.

(3) Heating the water from 0°C to 100°C:
Q: Why did the water start panicking at 100°C?
A: It realized it was boiling and going through some major changes, so it got a little nervous!

Again, we can use the formula Mass * C * ln(T2 / T1) to calculate the entropy change for this step.

(4) Evaporating the liquid water:
Q: Why did the water become a superstar?
A: It got tired of being "still" all the time, so it said, "I'm going to make a splash as steam!"

Now, for this step, we'll calculate the entropy change using the heat of vaporization divided by the corresponding temperature.

By adding up the individual entropy changes from each step, you'll get the total change in entropy from transforming your ice cube into steam. But remember, science is serious business, so make sure to double-check my results without relying too much on my jokes!

To calculate the change in entropy, we need to consider four different steps:

Step 1: Heating the ice from -12°C to 0°C.
Step 2: Melting the ice at 0°C.
Step 3: Heating the water from 0°C to 100°C.
Step 4: Evaporating the liquid water at 100°C.

To begin, let's calculate the entropy change for each step.

Step 1: Heating the ice from -12°C to 0°C.

The entropy change (ΔS1) for heating the ice can be calculated using the equation:
ΔS1 = mass * C * ln(T2 / T1)

Where:
mass = 27.9 g (mass of the ice cube)
C = specific heat capacity of ice = 2.09 J/g°C
T1 = initial temperature = -12°C
T2 = final temperature = 0°C

ΔS1 = 27.9 g * 2.09 J/g°C * ln(0°C / -12°C)

Note: The natural logarithm of a negative number is undefined, so we assume ln(-12°C) as -ln(12°C).

ΔS1 = 27.9 g * 2.09 J/g°C * -ln(12)

Step 2: Melting the ice at 0°C.

The entropy change (ΔS2) for melting the ice can be calculated using the equation:
ΔS2 = heat of fusion / T

Where:
heat of fusion = 334 J/g (entropy change required to convert ice to water at 0°C)
T = temperature = 0°C

ΔS2 = 334 J/g / 0°C

Note: Division by zero is undefined, so we can rewrite the equation as:
ΔS2 = limit as T approaches 0, 334 J/g / T

ΔS2 = 334 J/g / 0

Step 3: Heating the water from 0°C to 100°C.

The entropy change (ΔS3) for heating the water can be calculated using the equation:
ΔS3 = mass * C * ln(T2 / T1)

Where:
mass = 27.9 g (mass of the ice cube)
C = specific heat capacity of water = 4.18 J/g°C
T1 = initial temperature = 0°C
T2 = final temperature = 100°C

ΔS3 = 27.9 g * 4.18 J/g°C * ln(100°C / 0°C)

Note: The natural logarithm of zero is undefined, so we assume ln(0°C) as -ln(0.01°C) to avoid approaching zero.

ΔS3 = 27.9 g * 4.18 J/g°C * ln(100)

Step 4: Evaporating the liquid water at 100°C.

The entropy change (ΔS4) for evaporating the water can be calculated using the equation:
ΔS4 = heat of vaporization / T

Where:
heat of vaporization = 2260 J/g (entropy change required to convert water to steam at 100°C)
T = temperature = 100°C

ΔS4 = 2260 J/g / 100°C

ΔS4 = 22.6 J/g°C

Finally, to calculate the total change in entropy, we sum up the individual entropy changes for each step:
Total ΔS = ΔS1 + ΔS2 + ΔS3 + ΔS4

To calculate the change in entropy, we need to consider the entropy changes for each step involved in the transformation from a -12°C ice cube to steam at 115°C. Let's break down each step and calculate the entropy changes individually.

Step 1: Heating the ice from -12°C to 0°C
The entropy change for heating can be calculated using the equation: ΔS = m * C * ln(T2/T1), where ΔS is the change in entropy, m is the mass, C is the specific heat capacity, T2 is the final temperature, and T1 is the initial temperature.

Given:
Mass (m) of ice = 27.9 g
Specific heat capacity of ice (C) = 2.09 J/g°C
Initial temperature (T1) = -12°C
Final temperature (T2) = 0°C

Plugging in these values into the entropy change equation:
ΔS1 = 27.9 g * 2.09 J/g°C * ln(0°C/-12°C)

Step 2: Melting the ice
The entropy change for melting can be calculated by dividing the heat transferred (heat of fusion) by the temperature. The heat of fusion for water is 334 J/g.

Given:
Heat of fusion (Q) = 334 J/g

Plugging in this value into the entropy change equation:
ΔS2 = 27.9 g * (334 J/g / 0°C)

Step 3: Heating the water from 0°C to 100°C
Similar to Step 1, we can calculate the entropy change for heating using the formula: ΔS = m * C * ln(T2/T1).

Given:
Specific heat capacity of water (C) = 4.18 J/g°C
Initial temperature (T1) = 0°C
Final temperature (T2) = 100°C

Plugging in these values into the entropy change equation:
ΔS3 = 27.9 g * 4.18 J/g°C * ln(100°C/0°C)

Step 4: Evaporating the liquid water
The entropy change for vaporization can be calculated by dividing the heat transferred (heat of vaporization) by the temperature. The heat of vaporization for water is 40.7 kJ/mol, which can be converted to J/g by dividing by the molar mass of water (18.02 g/mol).

Given:
Heat of vaporization (Q) = 40.7 kJ/mol
Molar mass of water = 18.02 g/mol

Plugging in these values into the entropy change equation:
ΔS4 = 27.9 g * (40.7 kJ/mol / 18.02 g/mol)

Finally, to find the total change in entropy, we add up the entropy changes for each step:
ΔS_total = ΔS1 + ΔS2 + ΔS3 + ΔS4

Please plug in the values and perform the calculations to get the final change in entropy.

0.5×27.9×ln(273÷261)+(80×27.9/273)+ln(388/273)+(540/388)=10.54cal/k