The number of grams of ice that can be melted by 1 g of 100 c steam is?

Answer 1

- 80 calories are needed to melt 1 gram of 0°C ice to 1 gram of 0°C water

- 640 calories are released when 1 gram of steam at 100°C is condensed to water at 0°C

- 80/640 or 1/8 gram of steam at 100°C must be condensed

Answer 2

The latent heat condensation of steam is 540 calories. When each gram of steam condenses to water at 100 degrees Celsius water it releases this amount of energy. Thus

18 x 540c = 9720c

From there each gram of water releases 1 calorie for every degree of Celsius change in temperature.

18 x 99c = 1782c

These to combined bring the total amount of energy released to 11502 calories.

The latent heat of melting for water is 80 calories. From here we get

11502c / 80 = 143.775 grams of ice.

Answer 3

To determine the number of grams of ice melted by 1 g of 100°C steam, we use the heat equation: q = mcΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the temperature change. The heat required to melt ice (heat of fusion) is 334 J/g, so for 1 g of steam, it can melt 1g/334 J = 0.003 g of ice.

Answer 4

It depends on how 'cold' the ice is:

1 g of -1 0C needs less melting heat than 1 g of -20 0C

Answer 5

0.148