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Какое количество теплоты выделится за `t=10` с на ...

Creato il: 10 gennaio 2026

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Domanda

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Какое количество теплоты выделится за t=10 с на внутреннем сопротивлении аккумулятора E1{\mathcal{E}}_{1}, на внутреннем сопротивлении аккумулятора E2{\mathcal{E}}_{2} и резисторе R (по отдельности)? Схема цепи на рисунке.  E1=12{\mathcal{E}}_{1}=12 В,  E2=6{\mathcal{E}}_{2}=6 В, r_1=2 Ом, r_2=1 Ом, R=9 Ом.

Python

python
from PIL import Image, ImageEnhance
img=Image.open('/mnt/data/image.png')
img.size


(245, 294)

Python

python
import matplotlib.pyplot as plt, numpy as np
plt.figure(figsize=(4,5))
plt.imshow(img)
plt.axis('off')
plt.show()

Python

python
import numpy as np, matplotlib.pyplot as plt
arr=np.array(img)
# crop around left sources
crop=img.crop((0,40,120,250))
plt.figure(figsize=(4,6))
plt.imshow(crop)
plt.axis('off')
plt.show()

Pensiero

Analyzing circuit with opposing sources

In the diagram, the top cell has a longer upper plate, and the bottom cell has a shorter upper plate, making the positive terminal of the bottom cell at C. So, E1 and E2 are connected in opposition. Their EMFs oppose each other, so net EMF would be E1 - E2. Now, I need to figure out the current direction in this loop.

Calculating current and heat in the circuit

To find the current and heat in the resistors, I need to use the formula Q = I² * resistance * time. The total resistance for the series components is R_total = R + r1 + r2 = 9 + 2 + 1 = 12 ohms. Suppose we take the clockwise direction around the loop; we’ll apply KVL (Kirchhoff’s Voltage Law) to solve for the current. The net EMF is E1 - E2, resulting in an equation where E1 - E2 - I*R_total = 0.

Thought for 44s

Risposta

По схеме видно, что ЭДС включены встречно: у E1\mathcal{E}_1 «плюс» сверху (у узла AA), а у E2\mathcal{E}_2 «плюс» снизу (у узла CC). Поэтому результирующая ЭДС контура:

Eэкв=E1−E2=12−6=6 В.\mathcal{E}_\text{экв}=\mathcal{E}_1-\mathcal{E}_2=12-6=6\ \text{В}.

Полное сопротивление в цепи (все последовательно):

RΣ=R+r1+r2=9+2+1=12 Ω.R_\Sigma = R+r_1+r_2=9+2+1=12\ \Omega.

Ток:

I=EэквRΣ=612=0.5 А.I=\frac{\mathcal{E}_\text{экв}}{R_\Sigma}=\frac{6}{12}=0.5\ \text{А}.

Количество теплоты на каждом сопротивлении за t=10 сt=10\ \text{с}:

Q=I2 R t.Q = I^2\,R\,t.
  • На r1=2 Ωr_1=2\,\Omega:
Qr1=(0.5)2⋅2⋅10=5 Дж.Q_{r_1}=(0.5)^2\cdot 2\cdot 10=5\ \text{Дж}.
  • На r2=1 Ωr_2=1\,\Omega:
Qr2=(0.5)2⋅1⋅10=2.5 Дж.Q_{r_2}=(0.5)^2\cdot 1\cdot 10=2.5\ \text{Дж}.
  • На внешнем R=9 ΩR=9\,\Omega:
QR=(0.5)2⋅9⋅10=22.5 Дж.Q_{R}=(0.5)^2\cdot 9\cdot 10=22.5\ \text{Дж}.

Ответ: Qr1=5 ДжQ_{r_1}=5\ \text{Дж}, Qr2=2.5 ДжQ_{r_2}=2.5\ \text{Дж}, QR=22.5 ДжQ_{R}=22.5\ \text{Дж}.

Domanda

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решт

Pensiero

Solving thermodynamics problem for work done

I need to solve for the work done during the second expansion of an ideal monoatomic gas. We know the total work is 10 kJ, with some information about adiabatic and isobaric processes. I’ll use the ideal gas law and relations for these processes to link work in the two stages. T3 equals T1, so I need to find a way to connect the total work with the specific work done in each stage.

Connecting volumes and temperatures in thermodynamics

For the thermodynamics problem, I’m relating the state equations for both the adiabatic and isobaric processes. I have to connect temperatures and volumes with the ideal gas law. By working through equations like T2 = T1 (V1/V2)^(gamma-1) and others, I can relate state 2 and state 3 including their volume ratios.

Finalizing thermodynamics solution for work

I’ve established a relationship for V3 and P3, which aligns with the adiabatic and isobaric processes. Now the challenge is determining if the work done in the second stage (W23) is fixed or depends on parameter x. I also need to compute the work for both stages, either using pressure-volume changes or temperatures.

Final conclusion on the work calculation

I’ve calculated the work done in both the adiabatic (W12) and isobaric (W23) processes. It turns out the ratio of W12 to W23 is fixed at 3/2, and the total work is the sum of both. Thus, W23 is 4 kJ, which is 40% of the total work of 10 kJ.

Thought for 39s

Risposta

Для одноатомного идеального газа CV=32RC_V=\tfrac{3}{2}R.

1) Участок 1→21\to2 (адиабатическое расширение):
Q12=0⇒W12=−ΔU12Q_{12}=0\Rightarrow W_{12}=-\Delta U_{12}.

ΔU12=nCV(T2−T1)  ⇒  W12=nCV(T1−T2).\Delta U_{12}=nC_V(T_2-T_1)\;\Rightarrow\; W_{12}=nC_V(T_1-T_2).

2) Участок 2→32\to3 (изобарическое расширение):
Работа при p=constp=\text{const}:

W23=p(V3−V2)=nR(T3−T2).W_{23}=p(V_3-V_2)=nR(T_3-T_2).

По условию T3=T1T_3=T_1, значит

W23=nR(T1−T2).W_{23}=nR(T_1-T_2).

3) Связь работ:

W12W23=nCV(T1−T2)nR(T1−T2)=CVR=32.\frac{W_{12}}{W_{23}}=\frac{nC_V(T_1-T_2)}{nR(T_1-T_2)}=\frac{C_V}{R}=\frac{3}{2}.

Тогда

Wобщ=W12+W23=32W23+W23=52W23.W_{\text{общ}}=W_{12}+W_{23}=\frac{3}{2}W_{23}+W_{23}=\frac{5}{2}W_{23}.

Дано Wобщ=10 кДжW_{\text{общ}}=10\,\text{кДж}, значит

W23=25⋅10=4 кДж.W_{23}=\frac{2}{5}\cdot 10=4\,\text{кДж}.

Ответ: W2→3=4 кДж\boxed{W_{2\to3}=4\ \text{кДж}}.

Domanda

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реши

Pensiero

Analyzing thermodynamics process steps

I need to solve a thermodynamics problem for an ideal monatomic gas. First, process 1-2 is isothermal, since temperature is constant, and work done is 4 kJ. For process 2-3, since the line is straight and passes through the origin, it seems to represent an isobaric process.

Finding heat for isobaric process

The dashed line from the origin to point 2, and the solid line for process 2-3, suggest that 2-3 is isobaric. In an isobaric process, volume and temperature are proportional. Given that volume increased three times, temperature also tripled, and I calculate the total heat for process 2-3 to be about 20.8 kJ.

Thought for 19s

Risposta

Смотрим на график T(V)T(V):

  • участок 1→21\to2 — горизонтальный, значит изотерма (T=T= const);
  • участок 2→32\to3 — прямая, выходящая из начала координат (как показано пунктиром), значит изобара, потому что для идеального газа при p=p= const: T=pnR V⇒T∝V.T=\frac{p}{nR}\,V \quad\Rightarrow\quad T\propto V.

Дано: n=2n=2 моль, W12=4 кДжW_{12}=4\,\text{кДж}, в 2→32\to3: V3=3V2V_3=3V_2, T3=750 КT_3=750\,\text{К}.

1) Найдём температуры T1,T2T_1, T_2

На изобарe T∝VT\propto V, значит

T3T2=V3V2=3  ⇒  T2=7503=250 К.\frac{T_3}{T_2}=\frac{V_3}{V_2}=3 \;\Rightarrow\; T_2=\frac{750}{3}=250\ \text{К}.

А так как 1→21\to2 изотермический:

T1=T2=250 К.T_1=T_2=250\ \text{К}.

2) Теплота на участке 1→21\to2

Для изотермы идеального газа ΔU12=0\Delta U_{12}=0, поэтому

Q12=W12=4 кДж.Q_{12}=W_{12}=4\ \text{кДж}.

3) Теплота на участке 2→32\to3 (изобара)

Для одноатомного газа Cp=52RC_p=\frac{5}{2}R. Тогда

Q23=nCpΔT=2⋅52R (750−250)=5R⋅500=2500R.Q_{23}=nC_p\Delta T =2\cdot\frac{5}{2}R\,(750-250) =5R\cdot 500=2500R. Q23=2500⋅8.31≈2.08×104 Дж≈20.8 кДж.Q_{23}=2500\cdot 8.31 \approx 2.08\times 10^4\ \text{Дж}\approx 20.8\ \text{кДж}.

4) Итоговая теплота за процесс 1→2→31\to2\to3

Qобщ=Q12+Q23≈4+20.8=24.8 кДж.Q_{\text{общ}}=Q_{12}+Q_{23}\approx 4+20.8=24.8\ \text{кДж}.

Ответ: Q1−2−3≈24.8 кДж\boxed{Q_{1-2-3}\approx 24.8\ \text{кДж}}.

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