Answer: The volume of 10.4L will be expended
Explanation:
To calculate the volume of water that must be expended when the amount of energy released from the burning of 1.00L of crude oil.
The initial temperature of water is 18.5°C when temperature at steam is 285°C.
Since the water undergoes 3 stages:
Absorption of heat from 18.5°C to 100°C >> phase change to vapour at 100°C >> heating from 100°C to steam at 285°C
Given that:
Given Cw = 4186 J/kg°C
Cs = 2020 J/kg°C.
The latent heat of of vaporization of water Lv = 2.256 ×10⁶ J/kg
The total heat absorbed will be;
H = Cw × (100 – 18.5) + Lv + Cs × (285 – 100)
H = 4186×81.5 + 2.256×10⁶ + 2020× 185
H = 341159 + 2. 256 ×10⁶ + 373700
= 2.696 ×10⁶ J/kg
The amount of heat in J/L of water needed can be calculated as follows:
Density of water = 1000kg/m³ =
1000 kg/m³ × 1m³/1000L = 1kg/L (Basically conversion of density in kg/m³ to kg/L)
Let the volume of water needed be V litres.
Then the mass of water that must be expended = Density × Volume
= 1kg/L × V L = Vkg
The heat that would be absorbed by the water when 1L of crude oil is burned is V×H
= 2.696×10⁶ × V
This is also equal to 2.80×10⁷ J of energy (given).
So,
2.696×10⁶V = 2.8×10⁷
V = (2.80×10⁷)/(2.696×10⁶) = 10.4L of water.
