You have 40.00 g of ice at -20.0C. You want ot convert all the ice to steam at 140.0 C. How much energy will it take to do so?

q1 = heat needed to raise T of solid ice from -20 to 0.

q1 = mass ice x specific heat ice x (Tfinal-Tinitial).

q2 = heat needed to melt the solid ice at zero to liquid at zero.
q2 = mass ice x heat fusion.

q3 = heat needed to raise T of liquid water from zero C to 100 C.
q3 = mass H2O x specific heat liquid H2) x (Tfinal-Tinitial)

q4 = heat needed to convert liquid water at 100 C to steam at 100 C.
q4 = mass water x heat vaporization.

q5 = heat needed to raise T of steam from 100 C to 140 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial).
Total heat needed = q1 + q2 + q3 + q4 + q5.

Thanks DrBob222, I think I can figure this out now. I didn't know quite how to tackle this -- I appreciate it!

To find the amount of energy required to convert all the ice to steam, we need to consider the different phases of water and the energy required to change between them.

First, we need to heat the ice from its initial temperature (-20.0 °C) to its melting point (0.0 °C). The specific heat capacity of ice is 2.09 J/g·°C, which means it takes 2.09 Joules of energy to heat 1 gram of ice by 1 degree Celsius.

The temperature change required is: 0.0°C - (-20.0°C) = 20.0°C. Therefore, the energy required to heat the ice from -20.0 °C to 0.0 °C can be calculated using the equation:

Energy = mass × specific heat capacity × temperature change = 40.00 g × 2.09 J/g·°C × 20.0 °C

Next, we need to melt the ice. The heat of fusion for ice is 334 J/g, which means it takes 334 Joules of energy to convert 1 gram of ice at its melting point (0.0 °C) to liquid water at the same temperature.

To find the energy required to melt the ice, we multiply the mass of the ice by the heat of fusion:

Energy = mass × heat of fusion = 40.00 g × 334 J/g

After melting the ice, we need to heat the resulting water from 0.0 °C to 100.0 °C. The specific heat capacity of water is 4.18 J/g·°C.

The temperature change required is: 100.0 °C - 0.0 °C = 100.0 °C. Therefore, the energy required to heat the water from 0.0 °C to 100.0 °C can be calculated using the equation:

Energy = mass × specific heat capacity × temperature change = 40.00 g × 4.18 J/g·°C × 100.0 °C

Finally, we need to convert the water at 100.0 °C to steam at 100.0 °C. The heat of vaporization for water is 2260 J/g, which means it takes 2260 Joules of energy to convert 1 gram of water at its boiling point (100.0 °C) to steam at the same temperature.

To find the energy required to convert the water to steam, we multiply the mass of the water by the heat of vaporization:

Energy = mass × heat of vaporization = 40.00 g × 2260 J/g

To find the total energy required, we sum up the energies calculated for each step:

Total energy = Energy to heat the ice + Energy to melt the ice + Energy to heat the water + Energy to convert the water to steam

Once you plug in the values into the equations, you will get the total energy required in Joules.