A 478.18g ice cube is taken out of the freezer at -8.5 degrees Celsius. how much heat s needed to turn it in to steam?

q1 = heat needed to raise T from -8.5 to zero c.

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

q2 = heat needed to melt ice at Zero C and convert to liquid H2O.
q2 = mass ice x heat fusion

q3 = heat needed to raise T of water from zero C to 100 C

q3 = mass H2O x specific heat H2O x (Tfinal-Tinitial)_

q4 = heat needed to change liquid water at 100 C to steam at 100 C.
q4 = mass H2O x heat vaporization

Total = sum of individual qs.

To determine the amount of heat needed to turn the ice cube into steam, you'll need to consider the various phase changes involved in the process. Let's break it down step by step:

1. First, you need to heat the ice cube from its initial temperature (-8.5 degrees Celsius) to its melting point (0 degrees Celsius). The specific heat capacity of ice is approximately 2.09 J/g°C. So, the amount of heat required for this step can be calculated using the formula:

Q = m * C * ΔT

where Q is the amount of heat, m is the mass of the ice cube, C is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, we have:

Q1 = 478.18g * 2.09 J/g°C * (0 - (-8.5)°C)

2. Once the ice cube reaches the melting point, it undergoes a phase change from solid to liquid. This transition requires additional heat, known as the heat of fusion. The heat of fusion for water is approximately 334 J/g, which means that it takes 334 Joules of energy to convert each gram of ice at 0 degrees Celsius into water at 0 degrees Celsius. Therefore, the heat required for this step can be calculated as:

Q2 = m * ΔHf

where ΔHf is the heat of fusion. Plugging in the values, we have:

Q2 = 478.18g * 334 J/g

3. After the ice has completely melted, the water needs to be heated from 0 degrees Celsius to its boiling point (100 degrees Celsius). The specific heat capacity of water is approximately 4.18 J/g°C. Using the same formula as before, we can calculate the heat required for this step:

Q3 = m * C * ΔT

where Q3 is the amount of heat, m is the mass of water, C is the specific heat capacity, and ΔT is the change in temperature. Plugging in the values, we have:

Q3 = 478.18g * 4.18 J/g°C * (100 - 0)°C

4. Finally, the boiling water needs to be converted into steam. This phase change requires additional heat, known as the heat of vaporization. The heat of vaporization for water is approximately 2260 J/g. Therefore, the heat required for this step can be calculated as:

Q4 = m * ΔHvap

where ΔHvap is the heat of vaporization. Plugging in the values, we have:

Q4 = 478.18g * 2260 J/g

To find the total amount of heat required, we sum up the heat required for each step:

Total Heat = Q1 + Q2 + Q3 + Q4

Calculating this value will give you the amount of heat needed to turn the ice cube into steam.