Question

It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80×107 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 24.5 °C to 100 °C , it boils, and the resulting steam is raised to 325 °C.

Use 4186 J/(kg·°C) for the specific heat of water and 2020 J/(kg⋅°C) for the specific heat of steam.

amount of water in L= ?

Answer 1

Heat released by 1 L crude oil = 2.8*10 ⁷ J

So, heat absorbed by water = 2.8*10 ⁷ J

Heat absorbed by a substance when no phase change is occuring is given by m*s*(T ₂ - T ₁) where m is mass of the substance , s is its specific heat, T ₂= final temperature, T ₁= initial temperature.

Heat absorbed by a boiling substance is given by

m*L, where m is mass of given substance and L is latent heat of vaporisation.

Let the mass of water required be m kilograms.

For the temperature change from 24.5 to 100 degrees, heat absorbed = m* 4186* (100-24.5) = 316043 m Joules(here, we use s=4186 J/kg-C)

So,heat absorbed = 316.043 m KJ

For boiling of water, heat absorbed= m*L = 2260m kilojoules.

As, latent heat of vaporisation of water is 2260 KJ/kg

For temperature change of 100 degrees to 325 degrees of steam, heat absorbed is m*2020*(325-100)= 454500m Joules. (here,s=2020 KJ/kg-C)

So, heat absorbed= 454.500m KJ

So,net heat absorbed= 316.043m + 2260m + 454.5 m kilojoules= 3030.54m KJ

Also, heat absorbed by water is 2.8*10 ⁷ J = 2.8*10 ⁴ KJ

So, 3030.54m = 2.8* 10 ⁴ => m = 9.24 kg

Density of water is 1 kg/L

So, mass/volume=1 => mass = volume. So, volume required = 9.24 L