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Lena [83]
3 years ago
10

Which energy transformation occurs in an electric motor?

Chemistry
1 answer:
vivado [14]3 years ago
5 0

Answer:

Electrical energy to mechanical energy

Explanation:

In an electric motor, the energy transformation that is involve is a move from from electrical energy to mechanical energy.

For every system, according to the law of conservation of energy "energy is neither created nor destroyed but transformed from one form to another".

  • An electric motor converts electrical energy to mechanical energy
  • The electrical energy passes through and electromagnet set up.
  • This then cause the motion to rotate.
  • Therefore, electrical energy is transformed to mechanical energy.
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When 45 g of an alloy, at 25oC, are dropped into 100.0 g of water, the alloy absorbs 956 J of heat. If the final temperature of
taurus [48]

Answer:

The answer to the question is

The specific heat capacity of the alloy = 1.77 J/(g·°C)

Explanation:

To solve this, we list out the given variables thus

Mass of alloy = 45 g

Initial temperature of the alloy = 25 °C

Final temperature of the alloy = 37 °C

Heat absorbed by the alloy = 956 J

Thus we have

ΔH = m·c·(T₂ - T₁) where  ΔH = heat absorbed by the alloy = 956 J, c = specific heat capacity of the alloy and T₁ = Initial temperature of the alloy = 25 °C , T₂ = Final temperature of the alloy = 37 °C  and m = mass of the alloy = 45 g

∴ 956 J = 45 × C × (37 - 25) = 540 g·°C×c  or

c = 956 J/(540 g·°C) = 1.77 J/(g·°C)

The specific heat capacity of the alloy is 1.77 J/(g·°C)

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3 years ago
Tear off a small, flat sheet of waxed paper. Use the pipette to dispense a drop of water on the waxed paper. Now, dry the pipett
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Answer:

Explanation:

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A 1.2 L weather balloon on the ground has a temperature of 25°C and is at atmospheric pressure (1.0 atm). When it rises to an el
musickatia [10]

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Hello,

In this case, by using the general gas law, that allows us to understand the pressure-volume-temperature relationship as shown below:

\frac{P_1V_1}{T_1} =\frac{P_2V_2}{T_2}

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T_2=\frac{P_2V_2T_1}{P_1V_1}=\frac{1.8L*0.75atm*(25+273.15)K}{1.2L*1.0atm} \\\\T_2=335.42K=62.27^oC

Best regards.

4 0
3 years ago
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Does it take more, less, or the same amount of heat to melt 1.0 kg of ice at 0°C, or to bring 1.0 kg of liquid water at 0°C to t
Murljashka [212]

Answer : It takes less amount of heat to metal 1.0 Kg of ice.

Solution :

The process involved in this problem are :

(1):H_2O(s)(0^oC)\rightarrow H_2O(l)(0^oC)\\\\(2):H_2O(l)(0^oC)\rightarrow H_2O(l)(100^oC)

Now we have to calculate the amount of heat released or absorbed in both processes.

<u>For process 1 :</u>

Q_1=m\times \Delta H_{fusion}

where,

Q_1 = amount of heat absorbed = ?

m = mass of water or ice = 1.0 Kg

\Delta H_{fusion} = enthalpy change for fusion = 3.35\times 10^5J/Kg

Now put all the given values in Q_1, we get:

Q_1=1.0Kg\times 3.35\times 10^5J/Kg=3.35\times 10^5J

<u>For process 2 :</u>

Q_2=m\times c_{p,l}\times (T_{final}-T_{initial})

where,

Q_2 = amount of heat absorbed = ?

m = mass of water = 1.0 Kg

c_{p,l} = specific heat of liquid water = 4186J/Kg^oC

T_1 = initial temperature = 0^oC

T_2 = final temperature = 100^oC

Now put all the given values in Q_2, we get:

Q_2=1.0Kg\times 4186J/Kg^oC\times (100-0)^oC

Q_2=4.186\times 10^5J

From this we conclude that, Q_1 that means it takes less amount of heat to metal 1.0 Kg of ice.

Hence, the it takes less amount of heat to metal 1.0 Kg of ice.

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