To determine which substance has a latent heat of vaporization that matches the given energy required to boil 28.47 kg of the liquid, follow these steps:
### Step-by-Step Solution:
1. Given Information:
- Energy required to boil the liquid: [tex]\(10.14 \times 10^6 \; J\)[/tex]
- Mass of the liquid: [tex]\(28.47 \; kg\)[/tex]
- Latent heats of vaporization:
- Acetone: [tex]\(538,900 \; J/kg\)[/tex]
- Ammonia: [tex]\(1,371,000 \; J/kg\)[/tex]
- Propane: [tex]\(356,000 \; J/kg\)[/tex]
- Methane: [tex]\(480,600 \; J/kg\)[/tex]
- Ethanol: [tex]\(841,000 \; J/kg\)[/tex]
2. Calculate the energy required to boil 28.47 kg of each substance:
- Acetone:
[tex]\[
E_{\text{acetone}} = 538,900 \; \frac{J}{kg} \times 28.47 \; kg = 15,342,483 \; J
\][/tex]
- Ammonia:
[tex]\[
E_{\text{ammonia}} = 1,371,000 \; \frac{J}{kg} \times 28.47 \; kg = 39,032,370 \; J
\][/tex]
- Propane:
[tex]\[
E_{\text{propane}} = 356,000 \; \frac{J}{kg} \times 28.47 \; kg = 10,135,320 \; J
\][/tex]
- Methane:
[tex]\[
E_{\text{methane}} = 480,600 \; \frac{J}{kg} \times 28.47 \; kg = 13,682,682 \; J
\][/tex]
- Ethanol:
[tex]\[
E_{\text{ethanol}} = 841,000 \; \frac{J}{kg} \times 28.47 \; kg = 23,943,270 \; J
\][/tex]
3. Compare calculated energies with the given energy:
- The given energy required is [tex]\(10.14 \times 10^6 \; J\)[/tex].
Comparing this with the calculated energies:
- Acetone: [tex]\(15,342,483 \; J\)[/tex]
- Ammonia: [tex]\(39,032,370 \; J\)[/tex]
- Propane: [tex]\(10,135,320 \; J\)[/tex]
- Methane: [tex]\(13,682,682 \; J\)[/tex]
- Ethanol: [tex]\(23,943,270 \; J\)[/tex]
4. Find the closest match:
The energy required to boil propane ([tex]\(10,135,320 \; J\)[/tex]) is the closest to the given energy value of [tex]\(10.14 \times 10^6 \; J\)[/tex].
Therefore, the substance is Propane.
The correct choice is:
C. Propane
