Question

How many kilograms of water can be evaporated at 100 degrees Celsius by the combustion of 470 litres of fuel oil whose heat of combustion is 37 MJ/Litre

Answer 1

Answer on Question #64022, Physics / Molecular Physics | Thermodynamics

Question:

How many kilograms of water can be evaporated at 100 degrees Celsius by the combustion of 470 litres of fuel oil whose heat of combustion is 37 MJ/Litre?

Solution:

First, let's determine the heat generated during fuel's combustion:

$Q_{c} = V\cdot H_{c}$, where $V$ — fuel's volume, and $H_{c}$ — fuel's heat of combustion.

Now we must calculate the energy required for evaporation.

$Q_{e} = m\cdot H_{V}$, where $m$ is the mass of water, and $H_{V}$ — water's heat of vaporization.

$Q_{e} = Q_{c}$ that is $m\cdot H_{V} = V\cdot H_{c}$, and therefore $m = \frac{V\cdot H_{c}}{H_{V}}$.

$$V = 470\ \mathrm{L}$$ $$H_{c} = 37\ \mathrm{MJ/L} = 3.7\cdot 10^{7}\ \mathrm{J/L}$$ $$H_{V} = 2260\ \mathrm{KJ/kg} = 2.26\cdot 10^{3}\ \mathrm{J/kg}$$

Hence $m = \frac{470\cdot 3.7\cdot 10^{7}}{2.26\cdot 10^{3}} = 7.69\cdot 10^{6}\ \mathrm{kg}$

Answer:

$$7.69\cdot 10^{6}\ \mathrm{kg}$$

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