Energy Calculation: Propane Combustion Excitement!

How much mass of propane (in grams) is required to produce 547 kJ of energy when one pound of propane is combusted?

A) 10.9g

B) 11.2g

C) 12.5g

D) 9.8g

Answer:

The mass of propane (in grams) required to produce 547 kJ of energy is 10.9g.

Hello, energy enthusiasts!

When one pound of propane is combusted, it releases 22.8 MJ of energy. But how much mass of propane is needed to produce a specific amount of energy, like 547 kJ?

Using the conversion scale provided:

  • 1 pound = 453.592 g
  • 1 MJ = 10⁶ J
  • 1 KJ = 10³ J

Given that 22.8 MJ of energy is released per pound of propane, we can calculate the mass of propane needed to produce 547 kJ of energy:

547×10³ J = (547×10³ × 453.592) / 22.8×10⁶

Therefore, 547 kJ of energy requires 10.9g of propane to be combusted.

Voila! That's the exciting calculation of propane combustion energy. Stay tuned for more fascinating energy facts!

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