The Heat Transfer Process in a Water-Ice Mixture and Its Final Temperature

Let's explore the intriguing process of mixing a small amount of ice at -2°C with the same amount of boiling water (100°C) and discover the resulting mixture temperature. This scenario involves fundamental physics principles related to heat transfer, specific heat capacity, and latent heat of fusion.

Understanding the Problem

The specific heat capacity of ice is 2.108 kJ/kg, the specific heat capacity of water is 4.184 kJ/kg, and the latent heat of fusion of ice is 336 kJ/kg. The temperature of the ice is -2°C, and the temperature of the water is 100°C. When these two substances are mixed, the ice will first absorb heat to melt and then the resulting water will mix with the boiling water. The goal is to calculate the final temperature of the mixture.

Heat Transfer Analysis

First, we divide the process into three steps:

Step 1: Heat required to raise the temperature of the ice from -2°C to 0°C. Step 2: Heat required to melt the ice at 0°C. Step 3: Heat required to raise the temperature of the resulting water from 0°C to the final temperature ( T_C ).

Similarly, the heat released by the boiling water as it cools down to the final temperature ( T_C ) needs to be considered.

Step 1: Temperature Increase to 0°C

The heat required for the ice to change in temperature from -2°C to 0°C can be calculated using the formula:

Heat change ( m cdot c cdot Delta T )

Where ( m ) is the mass of the ice, ( c ) is the specific heat capacity of ice, and ( Delta T ) is the change in temperature.

For this step, the calculation is:

Heat absorbed by the ice 0.02 kg × 2.108 kJ/kg × [0 - (-2)] 0.08432 kJ

Step 2: Melting the Ice at 0°C

Melting the ice at 0°C requires the latent heat of fusion, which is given by:

Heat change ( m cdot L_f )

For this step, the calculation is:

Heat absorbed by the ice 0.02 kg × 336 kJ/kg 6.72 kJ

Step 3: Temperature Increase of Water from 0°C to ( T_C )

The heat required to raise the temperature of the ice-melted water to the final temperature can be calculated using:

Heat change ( m cdot c cdot Delta T )

For this step, the calculation is:

Heat absorbed by the water 0.02 kg × 4.184 kJ/kg × ( T_C )

Temperature Decrease of Boiling Water from 100°C to ( T_C )

The heat released by the boiling water as it cools to the final temperature can be calculated as:

Heat change ( m cdot c cdot Delta T )

For this step, the calculation is:

Heat released by the water 0.02 kg × 4.184 kJ/kg × (100 - ( T_C ))

Combining All the Steps

At thermal equilibrium, the total heat absorbed by the ice-melted water must equal the total heat released by the boiling water. Thus:

Heat absorbed by the ice-melted water Heat released by the boiling water

0.08432 kJ 6.72 kJ 0.02 kg × 4.184 kJ/kg × ( T_C ) 0.02 kg × 4.184 kJ/kg × (100 - ( T_C ))

Combining like terms:

8.368 kJ - 0.08368 ( T_C ) 8.368 kJ - 0.08368 ( T_C )

Solving for ( T_C ):

0.16736 ( T_C ) 1.56368

( T_C ) 9.3°C

Thus, the final temperature of the mixture is 9.3°C.

Conclusion

The final temperature of the mixture after the ice and water are combined is 9.3°C. This problem highlights the importance of understanding heat transfer processes and the relevant physical constants such as specific heat capacity and latent heat in thermodynamics.

Understanding these concepts is crucial for engineers, physicists, and anyone interested in the interesting mixture of different states of matter. If you have any further questions or want to explore other similar problems, feel free to ask!