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3 years ago

How much heat in joules would you need to raise the temperature of 1 kg of water by 5 °C?

1 answer:

7 0

First we need to know the specific heat capacity of water. You can look it up online because it is uniform. The specific heat capacity of water is 4184 kg^-1 °C^-1.

To calculate the energy required to heat a specific thing up, we can use the formula
E = mc△T
Where E is the total energy required, m as mass, c as specific heat capacity and △T

Substitute the numbers in
E = 1 x 4184 x 5
E = 20920J

Therefore the answer is 20920 joules

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what causes places near the equator to receive more direct sunlight than the poles? the poles are farther away from the sun than

Andru [333]

This is due to the tilt of the earth on its axis. Although the Sun shines on Earth, because of how the Earth is tilted, the equator is more directly hit compared to places found on the poles. The poles are hit at an angle, therefore the sunlight they receive is lesser than the places at the equator.

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3 years ago

A place that limits a wave’s motion.

Yuki888 [10]

Answer:

boundary

Explanation:

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2 years ago

What problems do you think might arise from not having a clear definition of "high crimes and misdemeanors"?

damaskus [11]

Answer:

People might misunderstand it and problems might happen because the audience will have different opinions.

Explanation:

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3 years ago

At the lowest point in a vertical dive (radius = 0.58 km), an airplane has a speed of 300 km/h which is not changing. Determine

murzikaleks [220]

Answer:

The centripetal acceleration is $a = 11.97 \ m/s^2$

Explanation:

From the question we are told that

The radius is $r = 0.58 \ km = 0.58 * 1000 = 580 \ m$

The speed is $v = 300\ km /hr = \frac{300 *1000}{1 * 3600 } = 83.33 \ m/s$

The centripetal acceleration of the pilot is mathematically represented as

$a = \frac{v^2 }{r}$

substituting values

$a = \frac{(83.33)^2 }{580}$

$a = 11.97 \ m/s^2$

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3 years ago

refrigerant 134a enters a compressor operating at steady state as saturated vapor at 0.12 MPa and exits at 1.2 MPa and 70 C at a

Afina-wow [57]

Answer:

the power input to the compressor is 7.19Kw

Explanation:

Hello!

To solve this problem follow the steps below.

1. We will call 1 the refrigerant state at the compressor inlet and 2 at the outlet.

2. We use thermodynamic tables to determine enthalpies in states 1 and 2.

(note: Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature. )

h1[quality=1, P=0.12Mpa)=237KJ/Kg

h2(P=1.2Mpa, t=70C)=300.6KJ/kg

3. uses the first law of thermodynamics in the compressor that states that the energy that enters a system is the same that must come out

Q=heat=0.32kJ/s

W=power input to the compressor

m=mass flow=0.108kg/S

m(h1)+W=Q+m(h2)

solving for W

W=Q+m(h2-h1)

W=0.32+0.108(300.6-237)=7.19Kw

the power input to the compressor is 7.19Kw

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3 years ago