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ValentinkaMS [17]
3 years ago
9

As a gas or liquid increases in heat, what direction will it naturally move?

Physics
1 answer:
miss Akunina [59]3 years ago
3 0

Hello!

Answer:

When a gas gets hot it should go up because of the pressure.

Explanation:

Hope this helps!

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A tray containing 0.20kg of water at 20degree celsius is placed in a freezer. The temperature of the water drops to 0degree cels
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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:

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