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.

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.

10.4L of water is expended when 1L of crude oil is burned.

Explanation:

This problem requires us to calculate the volume of water that must be expended to absorb the amount of energy released from the burning of 1.00L of crude oil.

In going from water at 18.5°C to steam at 285°C, 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 Cw = 4186 J/kg°C and Cs = 2020 J/kg°C. The latent heat of of vaporization of water Lv = 2.256 * 10⁶ J/kg

The total heat absorbed in the process per kilogram

H = Cw * (100 - 18.5) + Lv + Cs * (285 - 100)

H = 4186 * 81.5 + 2. 256 * 10⁶ + 2020 * 185

= 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.