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Vikki [24]
3 years ago
9

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, _____... is being

done.
A. COMPRESSION
B. WORK
C. RADIATION
Chemistry
2 answers:
Pachacha [2.7K]3 years ago
8 0

<u>Answer;</u>

-Work

If Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, <u>work </u> is being done.

<u>Explanation;</u>

  • Work refers to the application of a given force over a distance. Thus, we can say work is the product of force and distance.
  • Energy on the other hand is the ability of a body to change, its location, shape, or state of another body.
  • According to the work-Energy principle, a change in the kinetic energy, which is the energy possessed by a body in motion, is equivalent to the net work done on the body.
worty [1.4K]3 years ago
6 0

Answer: Option (B) is the correct answer.

Explanation:

Any activity which requires any physical or mental efforts in order to reach towards the result is known as work.

For example, when Jerome is swinging on a rope then he is doing a physical activity in which he is putting his efforts.

Thus, we can conclude that if Jerome is swinging on a rope and transferring energy from gravitational potential energy to kinetic energy, work is being done.

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Explanation:

The given data is as follows.

Volume of water = 0.25 m^{3}

Density of water = 1000 kg/m^{3}

Therefore,  mass of water = Density × Volume

                       = 1000 kg/m^{3} \times 0.25 m^{3}

                       = 250 kg  

Initial Temperature of water (T_{1}) = 20^{o}C

Final temperature of water = 140^{o}C

Heat of vaporization of water (dH_{v}) at 140^{o}C  is 2133 kJ/kg

Specific heat capacity of water = 4.184 kJ/kg/K

As 25% of water got evaporated at its boiling point (140^{o}C) in 60 min.

Therefore, amount of water evaporated = 0.25 × 250 (kg) = 62.5 kg

Heat required to evaporate = Amount of water evapotaed × Heat of vaporization

                           = 62.5 (kg) × 2133 (kJ/kg)

                           = 133.3 \times 10^{3} kJ

All this heat was supplied in 60 min = 60(min)  × 60(sec/min) = 3600 sec

Therefore, heat supplied per unit time = Heat required/time = \frac{133.3 \times 10^{3}kJ}{3600 s} = 37 kJ/s or kW

The power rating of electric heating element is 37 kW.

Hence, heat required to raise the temperature from 20^{o}C to 140^{o}C of 250 kg of water = Mass of water × specific heat capacity × (140 - 20)

                      = 250 (kg) × 40184 (kJ/kg/K) × (140 - 20) (K)

                     = 125520 kJ  

Time required = Heat required / Power rating

                       = \frac{125520}{37}

                       = 3392 sec

Time required to raise the temperature from 20^{o}C to 140^{o}C of 0.25 m^{3} water is calculated as follows.

                    \frac{3392 sec}{60 sec/min}

                     = 56 min

Thus, we can conclude that the time required to raise the temperature is 56 min.

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