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s2008m [1.1K]
3 years ago
5

Find the total amount of heat in Q lost through a wall 10' by 18' , with R value from q. 1. Inside temperature is 70 degrees F w

hile outside is 45 degrees F
Engineering
2 answers:
marissa [1.9K]3 years ago
6 0

Answer:

Just think

Explanation:

gogolik [260]3 years ago
5 0
That is really easy just think about it
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50.38
klemol [59]

Answer:

International Building Code (IBC)

Explanation:

6 0
2 years ago
An ideal gas initially at 300 K and 1 bar undergoes a three-step mechanically reversible cycle in a closed system. In step 12, p
Veseljchak [2.6K]

Answer:

Ts =Ta E)- 300(

569.5 K

5

Q12-W12 = -4014.26

Mol

AU2s = Q23= 5601.55

Mol

AUs¡ = Ws¡ = -5601.55

Explanation:

A clear details for the question is also attached.

(b) The P,V and T for state 1,2 and 3

P =1 bar Ti = 300 K and Vi from ideal gas Vi=

10

24.9x10 m

=

P-5 bar

Due to step 12 is isothermal: T1 = T2= 300 K and

VVi24.9 x 10x-4.9 x 10-3 *

The values at 3 calclated by Uing step 3l Adiabatic process

B-P ()

Since step 23 is Isochoric: Va =Vs= 4.99 m* and 7=

14

Ps-1x(4.99 x 103

P-1x(29x 10)

9.49 barr

And Ts =Ta E)- 300(

569.5 K

5

(c) For step 12: Isothermal, Since AT = 0 then AH12 = AU12 = 0 and

Work done for Isotermal process define as

8.314 x 300 In =4014.26

Wi2= RTi ln

mol

And fromn first law of thermodynamic

AU12= W12 +Q12

Q12-W12 = -4014.26

Mol

F'or step 23 Isochoric: AV = 0 Since volume change is zero W23= 0 and

Alls = Cp(L3-12)=5 x 8.311 (569.5 - 300) = 7812.18-

AU23= C (13-72) =5 x 8.314 (569.3 - 300) = 5601.53

Inol

Now from first law of thermodynamic the Q23

AU2s = Q23= 5601.55

Mol

For step 3-1 Adiabatic: Since in this process no heat transfer occur Q31= 0

and

AH

C,(T -Ts)=x 8.314 (300- 569.5)= -7842.18

mol

AU=C, (T¡-T)= x 8.314 (300

-5601.55

569.5)

mol

Now from first law of thermodynamie the Ws1

J

mol

AUs¡ = Ws¡ = -5601.55

3 0
3 years ago
What 2 forces move the secondary piston ahead?
jekas [21]

Answer:

The primary piston activates one of the two subsystems. The hydraulic pressure created, and the force of the primary piston spring, moves the secondary piston forward.

5 0
3 years ago
A 1 turn coil carries has a radius of 9.8 cm and a magnetic moment of 6.2 X 10 -2 Am 2. What is the current through the coil?
Alexus [3.1K]

Answer:

The current through the coil is 2.05 A

Explanation:

Given;

number of turns of the coil, N = 1

radius of the coil, r = 9.8 cm = 0.098 m

magnetic moment of the coil, P = 6.2 x 10⁻² A m²

The magnetic moment is given by;

P = IA

Where;

I is the current through the coil

A is area of the coil = πr² = π(0.098)² = 0.03018 m²

The current through the coil is given by;

I = P / A

I = (6.2 x 10⁻² ) / (0.03018)

I = 2.05 A

Therefore, the current through the coil is 2.05 A

6 0
3 years ago
Express the following quantities to the nearest standard prefix using no more than three digits.(a) 20,000,000 Hz(b) 1025 W(c) 0
bija089 [108]

Answer:

(a) 20 MHz

(b) 1.025 KW

(c) 3.33 ns

(d) 33 pF

Explanation:

(a) 20,000,000 Hz = 20 x 10^6 Hz = 20 Mega Hz = <u>20 MHz</u>

(b) 1025 W = 1.025 x 10^3 W = 1.025 Kilo W = <u>1.025 KW</u>

(c) 0.333 x 10^(-8) s = 3.33 x 10^(-9) s = 3.33 nano s = <u>3.33 ns</u>

(d) 33 x10^(-12)F = 33 pico F = <u>33 pF</u>

8 0
3 years ago
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