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mr Goodwill [35]
9 months ago
14

a hot pot of water is set on the counter to cool. after a few minutes it has lost 755 j of heat energy. how much heat energy has

the surrounding air gained?
Physics
1 answer:
velikii [3]9 months ago
5 0

The heat energy the surrounding air gained is <u>755 j.</u>

Energy is a quantitative property given to a body or physical system and can be identified as work being done in the form of heat and light. Energy is Conserved - The law of conservation of energy states that energy can be transformed into form, but cannot be created or destroyed. Energy is the ability to do work. It can exist in electrical, kinetic, thermal, electrical, chemical, nuclear, or other forms.

<u>Concept:-</u>

Since the energy is always conserved, the energy lost by the hot water is gained bt the surrounding

- energy lost  = + energy gained by the surrounding

- 755 j = + <u>755 j</u>

<u />

Energy exists in many forms. Examples include light energy, thermal energy, mechanical energy, gravitational energy, electrical energy, acoustic energy, chemical energy, and nuclear or atomic energy. Each form can be transformed or changed into other forms.

Learn more about energy here:-brainly.com/question/13881533

#SPJ1

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Answer:

a. Energy lost, Q = 16,800 Joules.

b. Power = 28 J/s

c. Time, t = 2357.14 seconds

d. I assumed that the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

Explanation:

<u>Given the following data;</u>

  • Mass = 0.20 kg
  • Initial temperature, T1 = 20°C
  • Final temperature = 0°C
  • Time = 10 minutes

a. To find the energy lost by the water as it cools to 0 degree celsius;

Mathematically, heat capacity is given by the formula;

Q = mcdt

Where;

  • Q represents the heat capacity or quantity of heat.
  • M represents the mass of an object.
  • C represents the specific heat capacity of water.
  • dt represents the change in temperature.

dt = T2 - T1

dt = 20 - 0

dt = 20°C

We know that the specific heat capacity of water is equal to  4200 J/kg°C

Substituting the values into the formula, we have;

Q = 0.20 * 4200 * 20

<em>Energy lost, Q = 16,800 Joules.</em>

b. To find the average rate at which the water is losing energy in J/s by using the following formula;

Power = \frac {energy}{time}

First of all, we would have to convert the value of time in minutes to seconds.

<u>Conversion:</u>

1 minute = 60 seconds

10 minutes = X seconds

Cross-multiplying, we have;

X = 60 * 10

X = 600 seconds

Substituting the values into the formula, we have;

Power = \frac {16800}{600}

<em>Power = 28 J/s</em>

c. To estimate the time taken for the water at 0 degree celsius to turn completely into ice;

We know that the latent heat of fusion of water is equal to 3.3 * 10⁵ J/kg.

Mathematically, the latent heat of fusion is calculated by using the formula;

Energy, Q = ml = pt

Substituting the values into the formula, we have;

0.20 * 3.3 * 10⁵ = 28 * t

0.20 * 330000 = 28t

66000 = 28t

t = \frac {66000}{28}

<em>Time, t = 2357.14 seconds.</em>

d. The assumption made is that, the ice remained at a temperature of zero degrees Celsius (0°C). Also, I assumed that the heat is being lost at a constant rate.

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Answer:

a = 4.96 m/s²

Explanation:

Given,

The mass of the box, m = 51 Kg

The magnitude of the applied force, Fₐ = 485 N

The friction force on the box, Fₓ = 232 N

The net force acting on the box is,

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Substituting the given values in the above equation

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