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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 40 L of crude oil, if the water has its temperature raised from 20.0∘C to 100∘C , it boils, and the resulting steam is raised to 300∘C.

You may need some or all of the following constants: The specific heat of water is 4186 J/kg ∘C and the specific heat of steam is 1520 J/kg ∘C. The heat of vaporization for water is 2256000 J/kg .

Respuesta :

Answer:

 m = 386.9 kg

Explanation:

This interesting exercise should use the calorimetry equations, for when we have temperature changes is

      Q= m ce ([tex]T_{f}[/tex] -T₀)

Heat for no temperature changes, but if state changes is

     Q = m L

For this problem the energy released by the oil is converted into heat in the water, let's calculate the energy released  

    Qc = m E  

    Qc = 40 2.80 10⁷  

    Qc = 1.12 10⁹ J  

This is the amount of heat that must be absorbed by the water, we separate in parts, Q1 is the heat to bring the water from 20ºC to 100ºC, Q2 is the heat to evaporate this water and Q3 is the heat to carry the steam from 100ºC to 300ºC , so the heat absorbed is  

     Qa = Q1 + Q2 + Q3  

      Q1= m ce ([tex]T_{f}[/tex] -T₀)

      Q1= m 4186 ( 100 -20.0)

      Q1= 3.349 10⁵ m

      Q2 = m L  

      Q2 = m 2,256 106  

      Q3= m ce ([tex]T_{f}[/tex] -T₀)

     Q3 = 3 1520 ( 300-100)

     Q3 = 3.04 10⁵ m

The total heat absorbed is

     Qa = 3,349 10⁵ m + 2,256 10⁶ m + 3.04 10⁵ m  

     Qa = m (3,349 10⁵ + 22.56 10⁵ + 3.04 10⁵)  

     Qa = m 28,949 10⁵  

As there are no losses  

     Qc = Qa  

     1.12 10⁹ = m 28.949 10⁵  

     m = 1.12 10⁹ / 28.949 10⁵  

     m= 3.86887 10² kg  

     m = 386.9 kg