In thermodynamics, heat is energy in transfer to or from a thermodynamic system, by mechanisms other than thermodynamic work or transfer of matter. The various mechanisms of energy transfer that define heat are stated in the next section of this article.
Like thermodynamic work, heat transfer is a process involving more than one system, not a property of any one system. In thermodynamics, energy transferred as heat contributes to change in the system's cardinal energy variable of state, for example its internal energy, or for example its enthalpy. This is to be distinguished from the ordinary language conception of heat as a property of an isolated system.
The quantity of energy transferred as heat in a process is the amount of transferred energy excluding any thermodynamic work that was done and any energy contained in matter transferred. For the precise definition of heat, it is necessary that it occur by a path that does not include transfer of matter.Though not immediately by the definition, but in special kinds of process, quantity of energy transferred as heat can be measured by its effect on the states of interacting bodies. For example, respectively in special circumstances, heat transfer can be measured by the amount of ice melted, or by change in temperature of a body in the surroundings of the system. Such methods are called calorimetry.
The conventional symbol used to represent the amount of heat transferred in a thermodynamic process is Q. As an amount of energy (being transferred), the SI unit of heat is the joule (J).
I tried the following:
We know, Q = mc (delta t)
but by Joule's law of heating, we also know that H = I^2 R t
Assuming that no heat is lost in the surroundings,
we get Q = H
=> mc (delta t) = I^2 R t
=> (delta t) = (I^2 R t)/(mc)
But m = V d = AL d
and R = (rho L) / A
=> (delta t) =...
Information online claims sun is 200 btu per sq ft in winter only when sun is perpendicular to plastic or glass of a green house. When sun hits glass or plastic at an angle there are losses. The steeper the angle the more losses will be.
Is it possible to calculate how warm a green house will...
My first problem is to find the absored and rejected heat. Can I say that it is equal to the work done in an isothermal proccess (##dQ=Pdv##)?
My reasoning : We have ##dQ=C_V d\theta + Pdv##. For constant temperature it becomes :$$dQ=Pdv$$
I guess the first one is wrong.
Because in this cycle we have ##|Q_H| > |Q_L|## then ## |Q_L| - |Q_H| ## is negative and caanot be equal with ##|W|##.
Am I right?
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I am new to this forum and I'm not even sure if this is the right place to ask but here goes:
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Consider a system which is characterized by the extensive variables ##(U,V,N_1,...,N_m)##. For a quasistatic process which occurs in contact with some pressure reservoir and where the ##N_i## are constant, one has
$$dU = TdS -PdV \implies TdS = dQ,$$
where the implication follows from the First...
I know process B absorbs heat but I can't determine the heat of process A.
In adiabatic process, Q = 0 but process A is not adiabatic. I only know both W and ΔU will be negative for process A but how to know Q?
Thanks
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I didn't have much trouble with part a but I'm struggling with b,c, and d. I considered the efficiency formula for a heat engine e = work done by engine/ qh
but i am unsure of how to approach it.
for part c) not sure how i can get to Tc without knowing Th
for d) my gut is telling me 5/2 but i...
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Energy lost by water = Energy gained by ice
Energy lost by water = 0.16 x 4200 x (100-t)
Energy gained by ice = 0.205 x L + 0.205 x (t) (where t is the temperature at thermal equilibrium). However, there does not appear to be enough info to continue.
The solution, however, considered t to be...
According to this,
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For this,
Dose anybody please know of a better way to derive the formula without having ##c = \frac{\Delta Q}{m \Delta T}## then taking the limit of both sides at ##\Delta T## approaches zero? I thought ##\Delta Q## like ##\Delta W## was not physically meaningful since by definition ##Q## is...
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For part(b),
My solution is,
##\Delta E_{int} = Q - W = \frac{3}{2}(P_fV_f - P_iV_i)##
##Q = W + \frac{3}{2}(P_fV_f - P_iV_i)##
##Q = 4000 + \frac{3}{2}((1 \times 10^6)(6 \times 10^{-3}) - (3 \times 10^6)(2 \times 10^{-3})##
##Q = 4000 J##
However, according to the solution b. ##−4000 J##
Can...
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Hello,
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Hi, zero education on heat transfer here, need some help.
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Does...
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https://www.nist.gov/system/files/documents/srd/jpcrd551.pdf for
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This would be for the transient conduction of a hollow cylinder, of wall...
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The properties are given below :
L = 5 mm
qin = 0
Tinf = 100 deg C
Tini = 20 deg C
rho = 7850 kg/m3
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k = 45.8 W/m.K
h = 20 W/m^2.K
alpha = k /...
Hello,
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Hello,
I am stuck how to proceed with the equations below.
The problem:
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Consider the following heat map:
from scipy import special
import numpy as np
import matplotlib.pyplot as plt
u0=200
r0x=25
r0y=25
rmax=2.5
alpha=2
t=0.575
y, x = np.meshgrid(np.linspace(0, 50, 100), np.linspace(0, 50, 100))
r=np.sqrt((x-r0x)**2+(y-r0y)**2)...