Respuesta :
Answer:
(a) 9.35 L (b) water will naturally settle at the bottom while oil will float on water which will affect how water extinguishes a fire.
Explanation:
(a) Using the equation below:
[tex]Q = m*c_{w}*(T_{2}-T_{1})+m*L+m*c_{v}*(T_{3}-T_{2})[/tex]
Where:
Q = energy in joules = 2.80*10^7 J
m = mass of water (kg)
[tex]c_{w}[/tex] = specific heat capacity of water = 4184 J/(kg*K)
[tex]c_{w}[/tex] = specific heat capacity of steam = 1996 J/(kg*K)
L = the latent heat of vaporization of water = 2260 kJ/kg
[tex]T_{1}[/tex] = 20 + 273.15 = 293.15 K
[tex]T_{2}[/tex] = 100 + 273.15 = 393.15 K
[tex]T_{3}[/tex] = 300 + 273.15 = 573.15 K
Therefore:
2.8*10^7 = m[4184(373.15-293.15)+2260000+ 1996(573.15-373.15)
2.8*10^7 = m[334720+2260000+399200]
2.8*10^7 = m[2993920]
Thus, m = 9.35 kg
However, mass = volume*density. And one liter of water has a mass of 1 kg.
Thus, the number of liters of water = 9.35 L
(b) Crude oil is less dense than water means the density of water is higher than that of crude oil. As a result of this, water will naturally settle at the bottom while oil will float on water which will affect how water extinguishes a fire.
a. The number of liters of water that must be expended to absorb the energy released by burning 1.00 Liter of crude oil is 9.35 Liters.
b. The density of crude oil is lesser than that of water because crude oil is less dense than water.
Consequently, water would always be at the bottom while crude oil would naturally float on water and as such limit the ability of water to extinguish a fire on a crude oil tanker.
Given the following data:
- Quantity of energy = [tex]2.80\times 10^7\;Joules[/tex]
- Volume of crude oil = 1.00 Liter.
- Initial temperature of water = 20.0°C
- Final temperature of water = 100.0°C
- Final temperature of steam = 300.0°C
Scientific data:
- Latent heat of vaporization of water = 2260 kJ/kg
- Specific heat capacity of water = 4184 J/kgK
- Density of water = 1000 [tex]kg/cm^3[/tex]
- Specific heat capacity of steam = 1996 J/kgK
Conversion:
Initial temperature of water = 20.0°C to K = [tex]273 + 20 = 293K[/tex]
Final temperature of water = 100.0°C = [tex]273 + 100 = 373K[/tex]
Final temperature of steam = 300.0°C = [tex]273 + 300 = 573K[/tex]
a. To calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 Liter of crude oil:
Mathematically, the quantity of heat produced during this process is given by the equation:
[tex]Q = M_wC_w(\theta_2-\theta_1) + M_wC_s(\theta_3-\theta_2) + M_wL[/tex]
Where:
- Q is the quantity of heat.
- [tex]M_w[/tex] is the mass of water.
- [tex]\theta[/tex] is the temperature.
- [tex]C_s[/tex] is the specific heat capacity of steam.
- [tex]C_w[/tex] is the specific heat capacity of water.
- L is the latent heat of vaporization of water.
Making mass the subject of formula, we have:
[tex]M_w = \frac{Q}{C_w(\theta_2-\theta_1) \;+ \;C_s(\theta_3-\theta_2)\;+\;L}[/tex]
Substituting the parameters into the formula, we have;
[tex]M_w = \frac{2.80\;\times \;10^7}{4184(373-293) \;+ \;1996(573-373)\;+\;2,260,000}\\\\M_w = \frac{2.80\;\times \;10^7}{4184(80) \;+ \;1996(200)\;+\;2,260,000}\\\\M_w = \frac{2.80\;\times \;10^7}{334,720 \;+ \;399,200\;+\;2,260,000}\\\\M_w = \frac{2.80\;\times \;10^7}{2,993,920}\\\\M_w = 9.35\;kg[/tex]
Mass of water = 9.35 kilograms.
By using direct proportion:
1 Liter of water = 1 kilogram of water
9.35 Liters of water = 9.35 kilograms of water
Therefore, the number of liters of water that must be expended to absorb the energy released by burning 1.00 Liter of crude oil is 9.35 Liters.
b. The density of crude oil is lesser than that of water because crude oil is less dense than water.
Consequently, water would always be at the bottom while crude oil would naturally float on water and as such limit the ability of water to extinguish a fire on a crude oil tanker.
Read more: https://brainly.com/question/18877825