Question

(a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80×107J 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’s temperature rises from 20.0°C to 100°C , it boils, and the resulting steam’s temperature rises to 300°C at constant pressure. (b) Discuss additional complications caused by the fact that crude oil is less dense than water.

Answer 1

Answer:

a) 9.22 L of water for 1 L of crude oil

Explanation:

a)The energy released by burning 1 liter of crude oil will raise the temperature of 'x' liters of the water from 20°C to 100°C, convert the liquid water into steam and raise the temperature of the steam to 300°C.

$Q_{crude oil} = 2.80 * 10^{7} J$

$C_{p,water}=4.2 *10^{3} J/kg.K$

$l_{steam}=2.3 *10^{6} J/kg$

$C_{p,steam}=2.0 *10^{3} J/kg.K$

$T_{1}=20\°C$

$T_{2}=100\°C$

$T_{3}=300\°C$

$m_{w}=? kg$

$Q_{crude oil} = m_{w}C_{p,water}(T_{2}-T_{1}) + m_{w}l_{steam} + m_{w}C_{p,steam}(T_{3}-T_{2})$

$2.80 * 10^{7} = m_{w}*4.2*10^{3}*(100-20) + m_{w}*2.3 *10^{6} + m_{w}*2.0 *10^{3}*(300-100)$

$2.80 * 10^{7} = 3.36*10^{5}m_{w} + 2.3 *10^{6}m_{w} + 4*10^{5}m_{w}$

$2.80 * 10^{7} = 3.036*10^{6}m_{w}$

$m_{w} = 9.22 kg$

$volume = \frac{mass}{density}$

$volume = \frac{9.22}{1}$

$volume = 9.22 L$

b) The lower density of crude oil compared to water causes the oil and water to separate with the oil on top. When the oil burns and releases heat which is absorbed by the water, the water vaporizes. This water vapor rises through the crude oil, taking particles of oil along with it. This means that the crude oil will spread through the air as it burns and vaporizes the water. This will further oxygenate the oil, increasing the rate of combustion and increasing the heat released. Adding water to an oil fire will increase the reaction rather than subduing it.

Answer 2

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:

$Q = m*c_{w}*(T_{2}-T_{1})+m*L+m*c_{v}*(T_{3}-T_{2})$

Where:

Q = energy in joules = 2.80*10^7 J

m = mass of water (kg)

$c_{w}$ = specific heat capacity of water = 4184 J/(kg*K)

$c_{w}$ = specific heat capacity of steam = 1996 J/(kg*K)

L = the latent heat of vaporization of water = 2260 kJ/kg

$T_{1}$  = 20 + 273.15 = 293.15 K

$T_{2}$  = 100 + 273.15 = 393.15 K

$T_{3}$ = 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.