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vlabodo [156]
3 years ago
14

7. En la sala de una casa hay una gran ventana de vidrio, por la que se presenta una pérdida significativa de calor; las medidas

de la ventana son 2,5 m x 1,5 m y 2,8 mm de espesor. Hallar la variación de calor en la unidad de tiempo si la temperatura exterior del vidrio es 16°C y la temperatura interior del vidrio es 20°C 8. De un cubo de 1 m de arista de un material X se saca de un horno a 40°C, se quiere saber qué cantidad de calor está radiando al exterior, si el coeficiente de emisividad (e) es 0,8 9. En un laboratorio hay una cantidad de 30 moles de un gas perfecto que sufre expansión isotérmica (temperatura constante). La presión inicial de esta masa de gas es de 20 atmosferas y el volumen es de 12 litros. Al final de la expansión, el volumen es de 50 litros. Por lo tanto, determine: a. La presión final de la masa de gas b. La temperatura que se produce en la transformación 10. Un sistema pasa de un estado a otro, intercambiando energía con el exterior. Calcular la variación de energía interna del sistema en los siguientes casos: a. El sistema libera 300 cal y realiza un trabajo de 700 J b. El sistema libera 300 cal y sobre él se realiza un trabajo de 500 J c. El sistema recibe 400 cal de calor y realiza un trabajo de 250 J
Physics
1 answer:
jonny [76]3 years ago
7 0

Answer:

I DON'T SPEAK TACO BELL!

Explanation:

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It took 1500 Newton's of force to push a car 3 meters. How much work was done
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3 0
2 years ago
3.25 kcal is the same amount of energy as
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3 years ago
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.1 An 8-ft 3 tank contains air at an initial temperature of 808F and initial pressure of 100 lbf/in. 2 The tank develops a small
Alina [70]

Correct temperature is 80°F

Answer:

T_f = 38.83°F

Explanation:

We are given;

Volume; V = 8 ft³

Initial Pressure; P_i = 100 lbf/in² = 100 × 12² lbf/ft²

Initial temperature; T_i = 80°F = 539.67 °R

Time for outlet flow; t_o = 90 s

Mass flow rate at outlet; m'_o = 0.03 lb/s

Final pressure; P_f = 30 lbf/in² = 30 × 12² lbf/ft²

Now, from ideal gas equation,

Pv = RT

Where v is initial specific volume

R is ideal gas constant = 53.33 ft.lbf/°R

Thus;

v = RT/P

v_i = 53.33 × 539.67/(100 × 12²)

v_i = 2 ft³/lb

Formula for initial mass is;

m_i = V/v_i

m_i = 8/2

m_i = 4 lb

Now change in mass is given as;

Δm = m'_o × t_o

Δm = 0.03 × 90

Δm = 2.7 lb

Now,

m_f = m_i - Δm

Thus; m_f = 4 - 2.7

m_f = 1.3 lb

Similarly in above;

v_f = V/m_f

v_f = 8/1.3

v_f = 6.154 ft³/lb

Again;

Pv = RT

Thus;

T_f = P_f•v_f/R

T_f = (30 × 12² × 6.154)/53.33

T_f = 498.5°R

Converting to °F gives;

T_f = 38.83°F

7 0
3 years ago
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