Isopropyl alcohol, C3H7OH is an organic compound. It melts at -89 C and boils at 82.6 C. IT has a heat fusion of 88 J/g and a heat of vaporization of 733.33 J/g. The specific heat of liquid isopropyl alcohol is 2.68 J/g-C. The specific heat of its solid form is 12.72 J/g-C. The specific heat of its gaseous form is 1.54 J/g-Cc. Determine the heat necessary to change 70 g of solid isopropyl alcohol at -120 C to gaseous isopropyl alcohol at 82.6 C
Here are the formulas you need for this problem. There are three.
When you are in the same phase (everything is liquid, or gases, or solids, you use
q = mass x specific heat x (Tfinal-Tinitiail). That's equation 1.
When you are at the phase change from solid to liquid use
q = mass x heat fusion. Let's call that equation 2.
When you are at the phase change from liquid to gas use
q = mass x heat vaporization. Let's call that equation 3.
For example, ice at -10 to steam at 120.
q1 = heat needed to move T from -10 as a solid to the melting point, but still a solid, is eqn 1 since you are in the same phase.
q2 = then you melt the solid at zero c to liquid at zero c. equn 2. This is a phase change.
q3 = heat needed to raise T of liquid water at zero C to liquid water at the boiling point of 100 C. Same phase. Eqn 1
q4 = heat needed to change liquid water at 100 to steam at 100. phase change. Eqn 3
q5 = heat needed to change steam at 100 to steam at 120. Same phase. eqn 1
Then total q is sum of q1 through q5.
Post your work if you get stuck.
To determine the heat necessary to change 70 g of solid isopropyl alcohol at -120°C to gaseous isopropyl alcohol at 82.6°C, we need to consider the energy required for each phase transition.
First, let's calculate the heat required to raise the temperature of the solid isopropyl alcohol from -120°C to its melting point of -89°C using the specific heat of the solid phase:
Q1 = mass × specific heat of solid × temperature change
= 70 g × 12.72 J/g-°C × (−89°C − (−120°C))
= 70 g × 12.72 J/g-°C × 31°C
= 27468 J
Next, we need to calculate the heat required to melt the solid isopropyl alcohol into liquid form:
Q2 = mass × heat fusion
= 70 g × 88 J/g
= 6160 J
Afterward, we need to calculate the heat required to raise the temperature of the liquid isopropyl alcohol from its melting point (-89°C) to its boiling point (82.6°C) using the specific heat of the liquid phase:
Q3 = mass × specific heat of liquid × temperature change
= 70 g × 2.68 J/g-°C × (82.6°C − (−89°C))
= 70 g × 2.68 J/g-°C × 171.6°C
= 31873.76 J
Lastly, we need to calculate the heat required to vaporize the liquid isopropyl alcohol into gaseous form:
Q4 = mass × heat of vaporization
= 70 g × 733.33 J/g
= 51333.1 J
To find the total heat required for the entire transformation, we sum up the individual heat values:
Total Heat = Q1 + Q2 + Q3 + Q4
= 27468 J + 6160 J + 31873.76 J + 51333.1 J
= 116834.86 J
Therefore, the total heat necessary to change 70 g of solid isopropyl alcohol at -120°C to gaseous isopropyl alcohol at 82.6°C is approximately 116834.86 J.
To determine the total heat necessary to change 70 g of solid isopropyl alcohol at -120°C to gaseous isopropyl alcohol at 82.6°C, we need to consider the various steps involved in the process and calculate the heat for each step separately.
Step 1: Heating the solid isopropyl alcohol from -120°C to its melting point.
We'll use the formula: Q = m * C * ΔT
where:
Q is the heat required (in joules),
m is the mass of the substance (in grams),
C is the specific heat of the solid form (in J/g-°C), and
ΔT is the change in temperature (in °C).
Q1 = 70 g * 12.72 J/g-°C * ((0°C) - (-120°C))
Q1 = 70 g * 12.72 J/g-°C * 120°C
Q1 = 106,848 J
Step 2: Melting the isopropyl alcohol at its melting point.
We'll use the formula: Q = m * ΔHf
where:
Q is the heat required (in joules),
m is the mass of the substance (in grams), and
ΔHf is the heat of fusion (in J/g).
Q2 = 70 g * 88 J/g
Q2 = 6,160 J
Step 3: Heating the liquid isopropyl alcohol from its melting point to its boiling point.
Q3 = 70 g * 2.68 J/g-°C * (82.6°C - 0°C)
Q3 = 15,368 J
Step 4: Vaporizing the liquid isopropyl alcohol at its boiling point.
Q4 = 70 g * 733.33 J/g
Q4 = 51,333.1 J
Step 5: Heating the gaseous isopropyl alcohol from its boiling point to 82.6°C.
Q5 = 70 g * 1.54 J/g-°C * (82.6°C - 0°C)
Q5 = 8,200.44 J
Total heat required = Q1 + Q2 + Q3 + Q4 + Q5
Total heat required = 106,848 J + 6,160 J + 15,368 J + 51,333.1 J + 8,200.44 J
Total heat required ≈ 188,909.54 J
Therefore, the heat necessary to change 70 g of solid isopropyl alcohol at -120°C to gaseous isopropyl alcohol at 82.6°C is approximately 188,909.54 J.