A 95.1-g ice cube is initially at 0.0°C.

(a) Find the change in entropy of the cube after it melts completely at 0.0°C. Hint: The latent heat of fusion for water is 3.33 105 J/kg. answer in J/kg

(b) What is the change in entropy of the environment in this process? answer in J/kg

Q=r•m

ΔS=Q/T₀=r•m/T₀=333•10⁵•0.0951/273=1.16•10⁴ J/K
If environment temperature is T,
ΔS= - Q/T= - 333•10⁵•0.0951/T= …

Oh, the melting ice cube dilemma! Let's tackle it with a bit of clowny humor!

(a) To find the change in entropy of the ice cube after it completely melts, we need to consider its initial and final states. In this case, the ice cube starts at 0.0°C and ends at 0.0°C as water. So, there's no change in temperature (ΔT = 0).

To calculate the change in entropy (ΔS), we can use the formula:

ΔS = Q / T

where Q is the heat absorbed (in this case, the latent heat of fusion for water) and T is the temperature in Kelvin.

Now, let's plug in the numbers. The latent heat of fusion for water is 3.33 x 10^5 J/kg, and since there's no change in temperature, T remains at 0°C or 273.15 K. So we have:

ΔS = (3.33 x 10^5 J/kg) / (273.15 K)

Solving that gives us the change in entropy of the ice cube after melting completely! Ta-da!

(b) Now, the change in entropy of the environment. Well, the ice cube has melted, so the environment gains heat. We can use the same formula:

ΔS = Q / T

Here, Q represents the heat lost by the ice cube. We can consider it equal to the latent heat of fusion for water since that's the amount of heat required to melt the ice cube. And T is still 273.15 K.

So, the change in entropy of the environment would be:

ΔS = (3.33 x 10^5 J/kg) / (273.15 K)

Voila! The change in entropy of the environment. Keep the jokes melting, my friend!

To solve this problem, we can apply the formula for entropy change:

ΔS = Q/T

where ΔS is the change in entropy, Q is the heat added or removed, and T is the temperature.

(a) To find the change in entropy of the ice cube after it melts completely, we need to calculate the heat required to melt the ice.
The heat required to melt the ice can be calculated using the formula:

Q = m * L

where Q is the heat, m is the mass of the ice, and L is the latent heat of fusion for water.

Given:
Mass of ice cube, m = 95.1 g
Latent heat of fusion for water, L = 3.33 * 10^5 J/kg

Converting the mass to kilograms:
m = 95.1 g = 0.0951 kg

Plugging the values into the formula:

Q = (0.0951 kg) * (3.33 * 10^5 J/kg)
Q = 3.16 * 10^4 J

Now, the change in entropy can be calculated as:

ΔS = Q/T

The temperature remains constant at 0.0°C, which is equivalent to 273.15 K.

ΔS = (3.16 * 10^4 J) / (273.15 K)
ΔS ≈ 115.72 J/K

Therefore, the change in entropy of the cube after it melts completely at 0.0°C is approximately 115.72 J/K.

(b) To find the change in entropy of the environment, we can use the fact that the overall change in entropy is zero for the whole process.

Therefore, the change in entropy of the environment would be the negative of the change in entropy of the ice cube:

ΔS_environment = -ΔS_cub

So, ΔS_environment ≈ -115.72 J/K

Hence, the change in entropy of the environment in this process is approximately -115.72 J/K.

To find the change in entropy of the ice cube as it melts completely, we can use the formula:

ΔS = Q / T

where ΔS is the change in entropy, Q is the heat transferred, and T is the temperature.

(a) Find the change in entropy of the cube after it melts completely at 0.0°C:

First, we need to find the heat transferred, which can be calculated using the formula:

Q = m * L

where Q is the heat transferred, m is the mass of the ice cube, and L is the latent heat of fusion for water.

Given: m = 95.1 g = 0.0951 kg (converting from grams to kilograms)
L = 3.33 * 10^5 J/kg

Plugging in the values, we get:

Q = 0.0951 kg * (3.33 * 10^5 J/kg) = 3.1593 * 10^4 J

Now, we need to find the change in entropy using the formula:

ΔS = Q / T

Since the temperature remains constant at 0.0°C, we can convert it to Kelvin by adding 273.15:

T = 0.0 + 273.15 = 273.15 K

Plugging in the values, we get:

ΔS = (3.1593 * 10^4 J) / (273.15 K) = 115.602 J/K

Therefore, the change in entropy of the cube after it melts completely at 0.0°C is 115.602 J/K.

(b) To find the change in entropy of the environment in this process, we can use the same formula:

ΔS = Q / T

In this case, the heat transferred is the same as before, which is 3.1593 * 10^4 J. However, the temperature of the environment remains constant at room temperature, which is typically around 298 K.

Plugging in the values, we get:

ΔS = (3.1593 * 10^4 J) / (298 K) = 105.993 J/K

Therefore, the change in entropy of the environment in this process is 105.993 J/K.