The Melting Ice in Warm Water: A Thermal Energy Analysis
When 0.03 kg of ice at 0°C is placed in a 0.02 kg of water at 50°C, the process is complex and involves several factors. This article will analyze the scenario, using thermodynamics to understand the outcome. We begin by examining the basic principles involved and then proceed to a detailed calculation.
Introduction to the Scenario
This particular scenario involves a common misconception: "The ice melts" is an oversimplified answer. This article will provide a more comprehensive explanation. Let us first consider the variables and the process involved.
Theoretical Background
The process of ice melting involves a phase change, where the latent heat of transformation is the key factor. The heat required to melt ice at 0°C is 80 cal/g. Here, we will also consider the specific heat capacity of water and ice, which are equal (Cice Cwater ≈ 1 cal/g°C). The latent heat of fusion for ice is 80 cal/g, and the specific heat capacity for water is 1 cal/g°C.
Calculation and Analysis
To determine the outcome, we need to compare the heat energy available in the warm water with the heat energy required to melt the ice.
Heat Energy Required to Melt the Ice
First, calculate the energy required to melt the ice:
E m × Lice-to-water
The mass of the ice is 0.03 kg (30 g), and the latent heat of melting ice is 80 cal/g:
E 30 g × 80 cal/g 2400 cal
Heat Energy Released by the Warm Water
Next, calculate the heat energy released by the warm water as it cools to 0°C:
E m × c × ΔT
The mass of the water is 0.02 kg (20 g), the specific heat capacity of water is 1 cal/g°C, and the temperature change is from 50°C to 0°C:
E 20 g × 1 cal/g°C × 50°C 1000 cal
Comparison of Heat Energy
Comparing the two sets of calories, we see that the warm water has only 1000 cal available, which is insufficient to melt 2400 cal of ice. Therefore, only part of the ice will melt, and the water will not reach 0°C.
Detailed Outcome
To determine the exact amount of ice that will melt, we use the heat energy left over:
Ice melted (1000 cal) / (80 cal/g) 12.5 g
This 12.5 g of ice will melt into water, and the remaining mass of water will be:
Remaining water 20 g - 12.5 g 17.5 g
The final state will be a mixture of 12.5 g of melted ice and 17.5 g of original warm water, with the total mass of the mixture being 30 g.
Conclusion
When 0.03 kg of ice at 0°C is placed in 0.02 kg of water at 50°C, the heat released by the warm water is insufficient to melt all the ice. The ice will partially melt, and the resulting mixture will be 12.5 g of liquid water and 17.5 g of ice in water equilibrium at 0°C.
Further Considerations
It is important to consider the volume of the ice and water. The volume of 0.03 kg of ice is greater than that of 0.02 kg of water, so the ice would not all fit into the container initially. The ice and water will eventually reach thermal equilibrium at 0°C, regardless of the ice's excess volume.