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kenny6666 [7]
4 years ago
6

When heat is converted into another form of energy, the total amount of energy is constant. Which law best illustrates this stat

ement?
the first law of thermodynamics
the second law of thermodynamics
the third law of thermodynamics
none of the above
Physics
2 answers:
Lorico [155]4 years ago
6 0
The correct option is (A) <span>the first law of thermodynamics

Explanation:
The first law of thermodynamics takes the idea of law of conservation of energy and modify it for thermodynamics systems. It is the total internal energy of the systems equals to the amount of heat added "to" the system and the workdone "by" the system given as:
</span>ΔU = Q - W
<span>Q = heat added TO the system
W = work BY the system
</span>ΔU = Total internal energy
<span>
But the total internal energy is CONSERVED; it means that energy cannot be created or destroyed; it can only be transformed from one form to another. Hence the correct option is (A).</span>
zalisa [80]4 years ago
5 0

Answer:

it is A

Explanation:

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What is one positive outcome of the recent<br>events?​
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Is position a base or derived quantity?
amid [387]

Position is measured in meters (m), so it is a base quantity.

<h3>What is base quantity?</h3>

A base or fundamental  quantity is a physical quantity, in which other quantities are derived from.

Example of fundamental quantities;

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8 0
2 years ago
What is the temperature 32 degrees Fahrenheit in degrees Celsius? A. 0°C B. 20°C C. 10°C D. –10°C
omeli [17]
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8 0
3 years ago
A mass of 5 kg of saturated water vapor at 150 kPa is heated at constant pressure until the temperature reaches 200°C. Calculate
yulyashka [42]

Answer:

The work done by the steam is 213 kJ.

Explanation:

Given that,

Mass = 5 kg

Pressure = 150 kPa

Temperature = 200°C

We need to calculate the specific volume

Using formula of work done

W=Pm\DeltaV

W=Pm(\dfrac{RT_{1}}{P_{atm}}-\dfrac{RT_{2}}{P_{atm}}

W=\dfrac{PmR}{P_{atm}}(T_{2}-T_{1})

Where,R = gas constant

T = temperature

P = pressure

P_{atm}=Atmosphere pressure

m = mass

Put the value into the formula

W=\dfrac{150\times10^{3}\times5\times287.05}{1.01\times10^{5}}\times(473-373)

W=213\ kJ

Hence, The work done by the steam is 213 kJ.    

6 0
3 years ago
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