Which formula can be used to solve problems related to the first law of thermodynamics? q = w u q = u – deltaw q = u w q = u – w

Physics

1 answer:

Orlov [11]2 years ago

3 0

"Q = ΔU + W" is the equation is used to solve the questions related to "First law of thermodynamics".

<h3> What is the first law of thermodyanamics?</h3>

"First law of thermodynamics" states that "energy" neither created nor destroyed, but it can transfer from "one form of energy" to "another form of energy".

This "First law of thermodynamics" is also called as "law of conversation of energy". The formula for "First law of thermodynamics" of a system is that "change in internal energy of a system" is same as the difference of "heat energy" flows across the " boundaries of a system" and the "work done" on the system.

ΔU = Q - W

Q = ΔU + W

Where, "ΔU" is "change in internal energy", "Q" is "heat transferred and "W" is "work done.

Hence "Q = ΔU + W" is the equation is used to solve the questions related to "First law of thermodynamics".

To know more about the First law of thermodynamics follow

brainly.com/question/15071682

You might be interested in

Two objects with masses of 2.75 kg and 4.45 kg are connected by a light string that passes over a light frictionless pulley to f

Llana [10]

Answer:

Explanation:

F = ma

4.45g - 2.75g = (4.45 + 2.75)a

a = 9.81(4.45 - 2.75) / (4.45 + 2.75) = 2.31625 ≈ 2.32 m/s²

T = 2.75(9.81 + 2.32) = 33.3 N

T = 4.45(9.81 - 2.32) = 33.3 N

b) 2.32 m/s² upward for 2.75 kg mass

2.32 m/s² downward for 4.45 kg mass

c) y = ½at² = ½(2.31625/3)1² = 1.158125 ≈ 1.16 m

4 0

2 years ago

A light platform is suspended from the ceiling by a spring. A student with a mass of 90 kg climbs onto the platform. When it sto

Ilya [14]

Refer to the diagram shown.

When the student climbs onto the platform, the spring stretches by 0.82 m to reach the equilibrium position.
The mass of the student is m = 90 kg, so his weight is
mg = (90 kg)*(9.8 m/s²) = 882 N

By definition, the spring constant is
k = (882 N)/(0.82 m) = 1075.6 N/m

When the spring is stretched by x from the equilibrium position, the restoring force is
F = - k*x.

If damping is ignored, the equation of motion is
F = m * acceleration
or
$m \frac{d^{2}x}{dt^{2}} = -kx \\ \frac{d^{2}x}{dt^{2}} + \frac{k}{m} x = 0$

Define ω² = k/m = 11.751 => ω = 3.457.
Then the solution of the ODE is
x(t) = c₁ cos(ωt) + c₂ sin(ωt)

x'(t) = -c₁ω sin(ωwt) + c₂ω cos(ωt)
When t=0, x' =0, therefore c₂ = 0

The solution is of the form
x(t) = c₁ cos(ωt)
When t = 0, x = 0.32 m. Therefore c₁ = 0.32

The motion is
x(t) = 0.32 cos(3.457t)
The single amplitude is 0.32 m, and the double amplitude is 0.64 m.

Answer:
0.32 m (single amplitude), or
0.64 m (double amplitude)