It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 × 10 7 J of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water has its temperature raised from 18.5 °C to 100 °C , it boils, and the resulting steam is raised to 285 °C. Use 4186 J/(kg⋅°C) for the specific heat of water and 2020 J/(kg⋅°C) for the specific heat of steam.

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.

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