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saveliy_v [14]
4 years ago
7

Electrical energy is used to turn the blades of a fan. The amount of energy transformed is seen here: 750 J electrical energy is

transformed into 400 J kinetic or mechanical energy. What happened to the remaining 350 J of energy? A) It was lost due to friction. B) It was transformed into 350 J of heat energy. C) It was destroyed by the rotating blades of the fan. D) It was never used and therefore was not transformed.
Chemistry
1 answer:
Alchen [17]4 years ago
6 0

B) It was transformed into 350 J of heat energy.

Explanation:

The remaining 350J of energy must have been transformed into 350J of heat energy.

A fan works by converting electrical energy into heat energy.

  • According to the third law of thermodynamics "no system is 100% efficient". The conversion of the energy from one form to another involves a lost in energy.
  • Heat is one very familiar way by which energy can be lost.
  • Some component energy is used to heat the fan in the process and it is a wasted energy.
  • Friction surfaces a heat energy in this process.

learn more:

Third law of thermodynamics brainly.com/question/3564634

#learnwithBrainly

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onsider the following reaction: CaCN2 + 3 H2O → CaCO3 + 2 NH3 105.0 g CaCN2 and 78.0 g H2O are reacted. Assuming 100% efficiency
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Answer : The excess reactant is, H_2O

The leftover amount of excess reagent is, 7.2 grams.

Solution : Given,

Mass of CaCN_2 = 105.0 g

Mass of H_2O = 78.0 g

Molar mass of CaCN_2 = 80.11 g/mole

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First we have to calculate the moles of CaCN_2 and H_2O.

\text{ Moles of }CaCN_2=\frac{\text{ Mass of }CaCN_2}{\text{ Molar mass of }CaCN_2}=\frac{105.0g}{80.11g/mole}=1.31moles

\text{ Moles of }H_2O=\frac{\text{ Mass of }H_2O}{\text{ Molar mass of }H_2O}=\frac{78.0g}{18g/mole}=4.33moles

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The balanced chemical reaction is,

CaCN_2+3H_2O\rightarrow CaCO_3+2NH_3

From the balanced reaction we conclude that

As, 1 mole of CaCN_2 react with 3 mole of H_2O

So, 1.31 moles of CaCN_2 react with 1.31\times 3=3.93 moles of H_2O

From this we conclude that, H_2O is an excess reagent because the given moles are greater than the required moles and CaCN_2 is a limiting reagent and it limits the formation of product.

Left moles of excess reactant = 4.33 - 3.93 = 0.4 moles

Now we have to calculate the mass of excess reactant.

\text{ Mass of excess reactant}=\text{ Moles of excess reactant}\times \text{ Molar mass of excess reactant}(H_2O)

\text{ Mass of excess reactant}=(0.4moles)\times (18g/mole)=7.2g

Thus, the leftover amount of excess reagent is, 7.2 grams.

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