The total amount of energy transferred as heat is equal to 288 Joules.
<u>Given the following data:</u>
- Internal energy = 123 Joules
To calculate the total amount of energy transferred as heat, we would apply the first law of thermodynamics.
<h3>The first law of thermodynamics.</h3>
Mathematically, the first law of thermodynamics is given by the formula:
<u>Where;</u>
is the change in internal energy.
- Q is the quantity of heat transferred.
Substituting the given parameters into the formula, we have;
Q = 288 Joules.
Read more on internal energy here: brainly.com/question/25737117
Answer:
7066kg/m³
Explanation:
The forces in these cases (air and water) are: Fa =mg =ρbVg Fw =(ρb −ρw)Vg where ρw = 1000 kg/m3 is density of water and ρb is density of the block and V is its density. We can find it from this two equations:
Fa /Fw = ρb / (ρb −ρw) ρb = ρw (Fa /Fa −Fw) =1000·(1* 21.2 /21.2 − 18.2)
= 7066kg/m³
Explanation:
1. Light Travelling from denser medium to less dense medium
2. Light Travelling from less dense medium to denser medium
Answer: m∠P ≈ 46,42°
because using the law of sines in ΔPQR
=> sin 75°/ 4 = sin P/3
so ur friend is wrong due to confusion between edges
+) we have: sin 75°/4 = sin P/3
=> sin P = sin 75°/4 . 3 = (3√6 + 3√2)/16
=> m∠P ≈ 46,42°
Explanation:
Work,
in thermodynamics, is the amount of energy that is transferred from one system
to another system without transfer of entropy. It is equal to the external
pressure multiplied by the change in volume of the system. It is expressed as
follows:<span>
W = PdV
Integrating and assuming that P is not affected
by changes in V or it is constant, then we will have:
W = P (V2 - V1)
Substituting the given values:
P = 1.0 atm = 101325 Pa
(V2 - V1) = 0.50 L =
W = 101325 N/m^3 ( 0.50) (1/1000) m^3
W = 50.66 N-m or 50.66 J
<span>
So, in the expansion process about 50.66 J of work is being done.
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