How much heat is required to melt 75.5g of ice at -25 degrees Celcius, and convert it to steam at 105 Celcius

To calculate the heat required to melt the ice and convert it to steam, we need to consider three steps individually:

1. Heat required to raise the temperature of the ice from -25 degrees Celsius to 0 degrees Celsius.
2. Heat required to melt the ice at 0 degrees Celsius.
3. Heat required to raise the temperature of the resulting water from 0 degrees Celsius to 100 degrees Celsius.
4. Heat required to convert the water at 100 degrees Celsius to steam at 105 degrees Celsius.

Let's calculate each step:

Step 1: Heat to raise the temperature of the ice from -25 degrees Celsius to 0 degrees Celsius.

The specific heat capacity of ice is 2.09 J/g°C.

Using the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we get:

Q1 = 75.5g * 2.09 J/g°C * (0°C - (-25°C))
= 75.5g * 2.09 J/g°C * 25°C
= 3934.875 J

Step 2: Heat required to melt the ice at 0 degrees Celsius.

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

Using the formula Q = m * ΔHf, where Q is the heat energy, m is the mass, and ΔHf is the heat of fusion, we get:

Q2 = 75.5g * 334 J/g
= 25127 J

Step 3: Heat to raise the temperature of the resulting water from 0 degrees Celsius to 100 degrees Celsius.

The specific heat capacity of water is 4.18 J/g°C.

Using the formula Q = m * c * ΔT, where Q is the heat energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature, we get:

Q3 = 75.5g * 4.18 J/g°C * (100°C - 0°C)
= 31477.5 J

Step 4: Heat required to convert the water at 100 degrees Celsius to steam at 105 degrees Celsius.

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

Using the formula Q = m * ΔHv, where Q is the heat energy, m is the mass, and ΔHv is the heat of vaporization, we get:

Q4 = 75.5g * 2260 J/g
= 170530 J

Now, adding up the heat required for each step:

Total heat required = Q1 + Q2 + Q3 + Q4
= 3934.875 J + 25127 J + 31477.5 J + 170530 J
= 229069.375 J

So, approximately 229,069.375 J of heat is required to melt 75.5g of ice at -25 degrees Celsius and convert it to steam at 105 degrees Celsius.

To determine how much heat is required to melt the ice and convert it to steam, we need to consider three phases: melting the ice to water, heating the water to its boiling point, and then vaporizing the water to steam.

Here's the step-by-step calculation:

1. Heat required to melt the ice:
The heat required to melt ice is given by the formula:
Q1 = m * ΔH1
where Q1 is the heat, m is the mass of ice, and ΔH1 is the heat of fusion of ice.

The heat of fusion of ice is 334 J/g.
Thus, Q1 = 75.5g * 334 J/g

2. Heat required to raise the temperature of the water to its boiling point:
The heat required to raise the temperature of water is given by the formula:
Q2 = m * C * ΔT
where Q2 is the heat, m is the mass of water, C is the specific heat capacity of water, and ΔT is the temperature change.

The specific heat capacity of water is approximately 4.18 J/g°C.
The temperature change is from -25°C to the boiling point of water at 100°C (ΔT = 100°C - (-25°C)).
Thus, Q2 = m * 4.18 J/g°C * (100°C - (-25°C))

3. Heat required to vaporize the water into steam:
The heat required to vaporize water is given by the formula:
Q3 = m * ΔH2
where Q3 is the heat, m is the mass of water, and ΔH2 is the heat of vaporization of water.

The heat of vaporization of water is 2260 J/g.
Thus, Q3 = 75.5g * 2260 J/g

4. Total heat required:
The total heat required is the sum of the heat required in each step:
Total heat = Q1 + Q2 + Q3

Now you can substitute the values into the calculations and find the total heat required to melt the ice and convert it to steam at the given temperatures.

45678j/c

q1=heat to move ice from -25 to zero C. (Not the correct spelling of celsius).

q1 = mass ice x specific heat ice x (Tfinal-Tinitial). Tfinal = zero and Tinitial = -25

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

q3 = heat to move water from zero C to 100 C.
q3 = mass x specific heat water x (Tfinal-Tinitial). Tfinal is 100 and Tinitial is zero.

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

q5 = heat to move steam from 100 C to 105 C.
q5 = mass steam x specific heat steam x (Tfinal-Tinitial)

Total heat required is
q1 + q2 + q3 + q4 + q5.
Post your work if you get stuck.