Answer:
The heat required to convert CCl3F from liquid at -50°C to gas at 50°C is 248, 296.64 J OR 248.30kJ
Explanation:
To calculate the heat required to convert CCl3F from a liquid at -50 °C to a gas at 50 °C, we calculate the heat required to increase the temperature of the liquid from -50°C to liquid at 0°C, then the heat needed to convert the liquid at 0°C to gas at 0°C, and then the heat needed to convert the gas at 0°C to gas at 50°C, adding these three values we obtain the total heat required.
Step 1: heat needed to increase the temperature from -50°C to 0°C
H = mcΔT ( m = 10g , C = 0.87 J/g, ΔT = ( B.p of CCl3F - (-50°C) = (23.8 +50) = 73.8°C
H = 10 * 0.87 * 73.8
H = 642.06 J
Step 2: heat required to convert liquid at 0°C to gas at °C
H = m Hv ( m = 10g, Hv = heat of vaporization = 24.75kJ/mol)
H = 10 * 24.75 *10^3
H = 247.5 *10^3 J
Step 3: heat needed to convert gas at 0°C oto gas at 50°C
H = mc T ( m =10g , c = 0.59J/g , T = (50°C- 23.8°C) = 26.2°C
H = 10 * 0.59 * 26.2
H = 154.58J
Step 4: adding the three values together
H = H1 + H2 + H3
H = 642.06 + 247500 + 154.58
H = 248, 296.64 J
h = 248.30kJ
Answer:
2.60 kJ
Explanation:
I have drawn a generalized heating curve for you below.
We are heating the liquid Freon to its boiling point, evaporating it completely, and then heating the vapour.
1. Heating the liquid
(a) Calculate ΔT
ΔT = [23.8 -(-50.0)] °C = (23.8 + 50.0) °C = 73.8 °C
(b) Calculate q₁
q₁ = mC₁ΔT = 10.0 × 0.87 × 73.8 = 642 J
2. Boiling the liquid
(a) Calculate the moles of Freon
[tex]n = \text{10.0 g} \times \dfrac{\text{1 mol}}{\text{137.37 g }} = \text{0.072 80 mol}[/tex]
(b) Calculate q₂
q₂ = nΔHvap = 0.07280 mol × 24 750 J/mol = 1802 J
3. Heating the vapour
(a) Calculate ΔT
ΔT = (50.0 - 23.8) °C = 26.2 °C
(b) Calculate q₃
q₃ = mC₃ΔT = 10.0 × 0.59 × 26.2 = 154 J
4. Calculate the total heat
q₁ = 642 J
q₂ = 1802 J
q₃ = 155 J
TOTAL = 2600 J = 2.60 kJ
It takes 2.60 kJ to heat the Freon from -50.0 °C to 50.0 °C .
