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IRISSAK [1]
2 years ago
5

A kettle consumes 198 kJ to boil water. If 168 kJ was used to boil the

Physics
1 answer:
Svetlanka [38]2 years ago
4 0

The efficiency of the kettle is 117.9%

<h3>What is efficiency?</h3>

Efficiency is used to measuring how much work or energy we can conserve in a process. In other words, it is like comparing the output of the energy to the input of the energy in any given system.

It is also define as the ratio of  Energy Input to the Energy Output. To express it in percent multiply it by 100.

Here, Efficiency = Energy output / Energy input

          η =\frac{W_{out} }{W_{in} } * 100

η = \frac{198kJ}{168kJ} * 100  

=  117.9%

So , the efficiency is 117.9%

For more Efficiency related question visit:

brainly.com/question/13110830

#SPJ1

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Answer:

1:00 pm

Explanation:

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FL is Eastern

pacific time is 3 hours less

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3 years ago
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What is a real world application of atwood's machine?
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3 years ago
The chemical reaction that causes iron to corrode in air is given by
sergiy2304 [10]

Answer :

(A) The value of \Delta G^o at 298 K is, 1521.9 kJ

(B) The value of \Delta G^o at 3652 K is, -301.59 kJ

Explanation :

Part A :

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = 1684 kJ = 1684000 J

\Delta S^o = standard entropy = 543.7 J/K

T = temperature of reaction = 298K

Now put all the given values in the above formula, we get:

\Delta G^o=(1684000J)-(298K\times 543.7J/K)

\Delta G^o=1521977.4J=1521.9kJ

Part B :

\Delta G^o=\Delta H^o-T\Delta S^o

where,

\Delta G^o = standard Gibbs free energy  = ?

\Delta H^o = standard enthalpy = 1684 kJ = 1684000 J

\Delta S^o = standard entropy = 543.7 J/K

T = temperature of reaction = 3652K

Now put all the given values in the above formula, we get:

\Delta G^o=(1684000J)-(3652K\times 543.7J/K)

\Delta G^o=-301592.4J=-301.59kJ

5 0
3 years ago
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Vadim26 [7]
Both waves would increase right? That seems correct since the water and air temp both equally changed.
4 0
3 years ago
A person walks 20.0° north of east for 3.20 km. How far would she have to walk due north and due east to arrive at the same loca
aivan3 [116]

Answer:

1.09 km, 3 km

Explanation:

displacement, d = 3.20 km at 20° North of east

A vector quantity has two components one along the x axis and the other is along Y axis. the component along X axis is called the horizontal component and it is due east.

The component along y axis is called vertical component and it is due North.

Displacement due North, dy = 3.20 Sin 20° = 1.09 km

Displacement due east, dx = 3.20 Cos 20° = 3 km

8 0
4 years ago
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