Show that 100 g of 100°C steam will completely melt
800 g of 0°C ice.
To show that 100 g of 100°C steam will completely melt 800 g of 0°C ice, we need to calculate and compare the amount of heat energy involved in the process.
The heat energy required to melt ice can be calculated using the latent heat of fusion, which is a characteristic property for a substance. For water, the latent heat of fusion is 334 J/g. This means that 334 Joules of heat energy is required to convert 1 gram of ice at 0°C into 1 gram of water at 0°C.
Therefore, the heat energy required to completely melt 800 g of ice can be calculated as follows:
Heat energy = mass of ice x latent heat of fusion
= 800 g x 334 J/g
= 267,200 J
Now, let's consider the heat energy released by steam when it condenses. The heat energy released during the condensation process can be calculated using the latent heat of vaporization. For water, the latent heat of vaporization is 2260 J/g. This means that 2260 Joules of heat energy are released when 1 gram of steam at 100°C condenses into 1 gram of water at 100°C.
So, if we have 100 g of steam, the heat energy released by its condensation can be calculated as follows:
Heat energy released = mass of steam x latent heat of vaporization
= 100 g x 2260 J/g
= 226,000 J
Comparing the heat energy required to melt the ice (267,200 J) with the heat energy released during the condensation of steam (226,000 J), we can see that the heat energy released by the steam is less than the heat energy required to melt the ice. Therefore, the 100 g of 100°C steam will not be able to completely melt the 800 g of 0°C ice.
To fully melt the ice, additional heat energy would be required.