Calculate the quantity of heat to convert 25.0g of ice at -10.0 degrees celcius to steam at 110.0 degrees celcius?

Note the correct spelling of celsius.

q1 = heat to move ice from -10 to zero.
q1 = mass ice x specific heat ice x (Tfinal-Tinitial) = 25 x ?? x [0-(-10)]

q2 = heat to melt the ice at zero C to liquid water at zero C.
q2 = mass ice x heat fusion.

q3 = heat to move liquid water at zero C to 100 C.
q3 = mass water x specific heat water x (Tfinal-Tinitial).

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

q5 = heat to move steam at 100 C to 110 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial).
Total Q = q1+q2+q3+q4+q5.

To calculate the quantity of heat required to convert a substance from one state to another, you need to consider the heat involved in each stage of the process.

First, calculate the heat required to raise the temperature of ice from -10.0 degrees Celsius to its melting point (0 degrees Celsius). This can be done using the specific heat formula:

q1 = m * C * ΔT

where:
q1 is the heat required (in joules),
m is the mass of the ice (in grams),
C is the specific heat capacity of ice (2.09 J/g°C),
ΔT is the change in temperature (in degrees Celsius).

q1 = 25.0g * 2.09 J/g°C * (0°C - (-10.0°C))
q1 = 25.0g * 2.09 J/g°C * 10.0°C
q1 = 522.5 J

The heat required to raise the temperature of the ice to its melting point is 522.5 joules.

Next, calculate the heat required to melt the ice at 0 degrees Celsius. The heat of fusion is the energy required to change a substance from a solid to a liquid without a change in temperature. For water, the heat of fusion is 334 J/g.

q2 = m * ΔHf

where:
q2 is the heat required (in joules),
m is the mass of the ice (in grams),
ΔHf is the heat of fusion of ice (334 J/g).

q2 = 25.0g * 334 J/g
q2 = 8,350 J

The heat required to melt the ice is 8,350 joules.

Next, calculate the heat required to raise the temperature of the water from 0 degrees Celsius to its boiling point (100 degrees Celsius). This can be done using the specific heat formula:

q3 = m * C * ΔT

where:
q3 is the heat required (in joules),
m is the mass of the water (in grams),
C is the specific heat capacity of water (4.18 J/g°C),
ΔT is the change in temperature (in degrees Celsius).

q3 = 25.0g * 4.18 J/g°C * (100.0°C - 0°C)
q3 = 25.0g * 4.18 J/g°C * 100.0°C
q3 = 10,450 J

The heat required to raise the temperature of the water to its boiling point is 10,450 joules.

Finally, calculate the heat required to convert the water at 100 degrees Celsius to steam at 110 degrees Celsius. The heat of vaporization is the energy required to change a substance from a liquid to a gas without a change in temperature. For water, the heat of vaporization is 2,260 J/g.

q4 = m * ΔHv

where:
q4 is the heat required (in joules),
m is the mass of the water (in grams),
ΔHv is the heat of vaporization of water (2,260 J/g).

q4 = 25.0g * 2,260 J/g
q4 = 56,500 J

The heat required to convert water to steam is 56,500 joules.

To find the total quantity of heat required, sum up all the individual heat values:

Total heat = q1 + q2 + q3 + q4
Total heat = 522.5 J + 8,350 J + 10,450 J + 56,500 J
Total heat = 75,822.5 J

Therefore, the quantity of heat required to convert 25.0g of ice at -10.0 degrees Celsius to steam at 110.0 degrees Celsius is approximately 75,822.5 joules.

To calculate the quantity of heat needed to convert a substance from one phase to another, we need to consider the following steps:

Step 1: Heating the ice from -10.0 °C to 0 °C.
Step 2: Melting the ice at 0 °C.
Step 3: Heating the water from 0 °C to 100 °C.
Step 4: Vaporizing the water at 100 °C.
Step 5: Heating the steam from 100 °C to 110 °C.

Let's calculate the heat required for each step:

Step 1: Heating the ice from -10.0 °C to 0.0 °C.
To heat a substance, we use the specific heat capacity formula:
q = m * c * ΔT,
where q is the heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

The specific heat capacity of ice is 2.093 J/g°C.

ΔT = 0 °C - (-10.0 °C) = 10.0 °C.
q1 = m * c * ΔT = 25.0 g * 2.093 J/g°C * 10.0 °C = 522.5 J.

Step 2: Melting the ice at 0.0 °C.
The heat required to melt a substance is calculated using the formula:
q = m * ΔHf,
where q is the heat, m is the mass, and ΔHf is the heat of fusion.

The heat of fusion for ice is 334 J/g.

q2 = m * ΔHf = 25.0 g * 334 J/g = 8,350 J.

Step 3: Heating the water from 0.0 °C to 100.0 °C.
Using the specific heat capacity formula:
q = m * c * ΔT.

The specific heat capacity of water is 4.184 J/g°C.

ΔT = 100.0 °C - 0.0 °C = 100.0 °C.
q3 = m * c * ΔT = 25.0 g * 4.184 J/g°C * 100.0 °C = 10,460 J.

Step 4: Vaporizing the water at 100.0 °C.
Using the formula:
q = m * ΔHv,
where q is the heat, m is the mass, and ΔHv is the heat of vaporization.

The heat of vaporization for water is 2260 J/g.

q4 = m * ΔHv = 25.0 g * 2260 J/g = 56,500 J.

Step 5: Heating the steam from 100.0 °C to 110.0 °C.
Using the specific heat capacity formula:
q = m * c * ΔT.

The specific heat capacity of steam is 2.03 J/g°C.

ΔT = 110.0 °C - 100.0 °C = 10.0 °C.
q5 = m * c * ΔT = 25.0 g * 2.03 J/g°C * 10.0 °C = 507.5 J.

Now, we can calculate the total quantity of heat required to convert 25.0 g of ice at -10.0 °C to steam at 110.0 °C by summing up the heat for each step:

Total heat = q1 + q2 + q3 + q4 + q5
Total heat = 522.5 J + 8,350 J + 10,460 J + 56,500 J + 507.5 J
Total heat = 76,340 J

Therefore, the quantity of heat required is 76,340 J.