How much heat is required to melt 50 grams of ice heat it to its boiling point and then turn it into vapor

To determine the amount of heat required to melt 50 grams of ice, heat it to its boiling point, and then turn it into vapor, we need to consider the different phases and the specific heat capacities of each phase.

First, let's calculate the heat required to melt the ice:

1. Heat required to melt the ice (from solid to liquid):
The heat required can be calculated using the formula:
Q = m * ΔHf
where Q is the heat, m is the mass of the substance, and ΔHf is the heat of fusion (latent heat) for ice, which is 334 J/g.

Q = 50 g * 334 J/g = 16700 J

Next, let's calculate the heat required to raise the temperature of water from its melting point to its boiling point:

2. Heat required to raise the temperature of water:
The heat required can be calculated using the formula:
Q = m * c * ΔT
where Q is the heat, m is the mass of the substance, c is the specific heat capacity of water, and ΔT is the change in temperature.

The specific heat capacity of water is approximately 4.18 J/g°C.
The temperature change required is from 0°C (melting point of ice) to 100°C (boiling point of water).

Q = 50 g * 4.18 J/g°C * 100°C = 20900 J

Finally, let's calculate the heat required to vaporize the water:

3. Heat required to vaporize the water (from liquid to vapor):
The heat required can be calculated using the formula:
Q = m * ΔHv
where Q is the heat, m is the mass of the substance, and ΔHv is the heat of vaporization (latent heat) for water, which is 2260 J/g.

Q = 50 g * 2260 J/g = 113000 J

Now, add up the heat required for each phase change:

Total heat required = Heat to melt ice + Heat to raise temperature + Heat to vaporize water
Total heat required = 16700 J + 20900 J + 113000 J
Total heat required = 150600 J

Therefore, it would take 150,600 joules of heat to melt 50 grams of ice, heat it to its boiling point, and then turn it into vapor.