60 dkg water with 70 °C is mixed with 10 dkg ice with -10 °C. How much will the temperature of the mixed material, and how change the entropy of the system? Prove that this process is irreversible!

To solve the problem of mixing 60 dkg of water at 70 °C with 10 dkg of ice at -10 °C, we need to consider the principles of heat transfer and the concept of entropy. Let's break this down step by step.

Calculating the Final Temperature

First, we need to determine how much heat is exchanged between the water and the ice. The water will lose heat as it cools down, while the ice will gain heat as it warms up and eventually melts. We can use the specific heat capacities and the latent heat of fusion for ice in our calculations.

Specific Heat Capacities

• Specific heat of water, \( c_w = 4.18 \, \text{J/g°C} \)
• Specific heat of ice, \( c_i = 2.09 \, \text{J/g°C} \)
• Latent heat of fusion for ice, \( L_f = 334 \, \text{J/g} \)

Heat Transfer Calculations

1. **Heating the ice from -10 °C to 0 °C**:

To warm the ice, we use the formula:

Q_1 = m_i \cdot c_i \cdot \Delta T

Where:

• m_i = 10 dkg = 1000 g
• ΔT = 0 - (-10) = 10 °C

Calculating Q_1:

Q_1 = 1000 g * 2.09 J/g°C * 10 °C = 20,900 J

2. **Melting the ice at 0 °C**:

Next, we calculate the heat required to melt the ice:

Q_2 = m_i \cdot L_f

Q_2 = 1000 g * 334 J/g = 334,000 J

3. **Cooling the water from 70 °C to the final temperature (T_f)**:

Let’s denote the final temperature as T_f. The heat lost by the water can be expressed as:

Q_3 = m_w \cdot c_w \cdot (T_w - T_f)

Where:

• m_w = 60 dkg = 6000 g
• T_w = 70 °C

So, Q_3 = 6000 g * 4.18 J/g°C * (70 - T_f)

Setting Up the Equation

Now, we can set up the energy balance equation:

Q_3 = Q_1 + Q_2

6000 g * 4.18 J/g°C * (70 - T_f) = 20,900 J + 334,000 J

25,080 (70 - T_f) = 354,900

70 - T_f = 354,900 / 25,080

70 - T_f ≈ 14.14

T_f ≈ 70 - 14.14 ≈ 55.86 °C

Entropy Change of the System

Next, we need to calculate the change in entropy for both the water and the ice. The total change in entropy (ΔS) is the sum of the entropy changes of the water and the ice.

Entropy Change for Water

ΔS_w = m_w * c_w * ln(T_f / T_w)

ΔS_w = 6000 g * 4.18 J/g°C * ln((55.86 + 273) / (70 + 273))

ΔS_w ≈ 6000 * 4.18 * ln(328.86 / 343) ≈ -1,000 J/K

Entropy Change for Ice

For the ice, we have two parts: warming the ice to 0 °C and melting it.

1. **Heating the ice**:

ΔS_i1 = m_i * c_i * ln(T_f / T_i)

ΔS_i1 = 1000 g * 2.09 J/g°C * ln(273 / 263) ≈ 1,000 J/K

2. **Melting the ice**:

ΔS_i2 = m_i * L_f / T_m

ΔS_i2 = 1000 g * 334 J/g / 273 K ≈ 1,224 J/K

So, the total entropy change for the ice is:

ΔS_i = ΔS_i1 + ΔS_i2 ≈ 1,000 + 1,224 ≈ 2,224 J/K

Total Entropy Change

Now, we can find the total change in entropy:

ΔS_total = ΔS_w + ΔS_i ≈ -1,000 + 2,224 ≈ 1,224 J/K

Irreversibility of the Process

A process is considered irreversible if the total entropy change of the system is positive. In this case, we found that the total entropy change is approximately 1,224 J/K, which is positive. This indicates that the mixing of hot water and cold ice leads to a more disordered state, confirming the irreversibility of the process.

In summary, the final temperature of the mixture is about 55.86 °C, and the total entropy change of the system is positive, demonstrating that the process is indeed irreversible. This aligns with the second law of thermodynamics, which states that the entropy of an isolated system always increases over time.