ITS A TESTS PLSSS I NEED THIS RN

During an office party, an office worker claims that a can of cold beer on his table warmed up to 20oC by picking up energy from

Contact [7]

Answer:

Yes

Explanation:

As we know that heat transfer take place from high temperature body to low temperature body.

In the given problem ,the temperature of the air is high as compare to the temperature of can of bear ,so the heat transfer will take place from air to can of bear and at the last stage when temperature of can of bear will become to the temperature of air then heat transfer will be stop.Because temperature of the both body will become at the same and this stage is called thermal equilibrium.

So an office worker claim is correct.

4 0

3 years ago

An open vat in a food processing plant contains 500 L of water at 20°C and atmospheric pressure. If the water is heated to 80°C,

tester [92]

Answer:

percentage change in volume is 2.60%

water level rise is 4.138 mm

Explanation:

given data

volume of water V = 500 L

temperature T1 = 20°C

temperature T2 = 80°C

vat diameter = 2 m

to find out

percentage change in volume and how much water level rise

solution

we will apply here bulk modulus equation that is ratio of change in pressure to rate of change of volume to change of pressure

and we know that is also in term of change in density also

so

E = $-\frac{dp}{dV/V}$ ................1

And $-\frac{dV}{V} = \frac{d\rho}{\rho}$ ............2

here ρ is density

and we know ρ for 20°C = 998 kg/m³

and ρ for 80°C = 972 kg/m³

so from equation 2 put all value

$-\frac{dV}{V} = \frac{d\rho}{\rho}$

$-\frac{dV}{500*10^{-3} } = \frac{972-998}{998}$

dV = 0.0130 m³

so now % change in volume will be

dV % = $-\frac{dV}{V}$ × 100

dV % = $-\frac{0.0130}{500*10^{-3} }$ × 100

dV % = 2.60 %

so percentage change in volume is 2.60%

and

initial volume v1 = $\frac{\pi }{4} *d^2*l(i)$ ................3

final volume v2 = $\frac{\pi }{4} *d^2*l(f)$ ................4

now from equation 3 and 4 , subtract v1 by v2

v2 - v1 = $\frac{\pi }{4} *d^2*(l(f)-l(i))$

dV = $\frac{\pi }{4} *d^2*dl$

put here all value

0.0130 = $\frac{\pi }{4} *2^2*dl$

dl = 0.004138 m

so water level rise is 4.138 mm

8 0

3 years ago

A piston-cylinder assembly has initially a volume of 0.3 m3 of air at 25 °C. Mass of the air is 1 kg. Weights are put on the pis

kogti [31]

Answer:

n=2.32

w= -213.9 KW

Explanation:

$V_1=0.3m^3,T_1=298 K$

$V_2=0.1m^3,T_1=1273 K$

Mass of air=1 kg

For polytropic process $pv^n=C$ ,n is the polytropic constant.

$Tv^{n-1}=C$

$T_1v^{n-1}_1=T_2v^{n-1}_2$

$298\times .3^{n-1}_1=1273\times .1^{n-1}_2$

n=2.32

Work in polytropic process given as

w= $\dfrac{P_1V_1-P_2V_2}{n-1}$

w= $mR\dfrac{T_1-T_2}{n-1}$

Now by putting the values

w= $1\times 0.287\dfrac{289-1273}{2.32-1}$

w= -213.9 KW

Negative sign indicates that work is given to the system or work is done on the system.

For T_V diagram

We can easily observe that when piston cylinder reach on new position then volume reduces and temperature increases,so we can say that this is compression process.

5 0

3 years ago

A coal fired power plant geneartes 2.4 lbs. of CO2 per kWh. A lighting system consumes 300,000kWh per year. A corporation is con

Serjik [45]

Answer:

The perceived economic impact of CO2 generated per year by lighting sstem is $8164.67.

Explanation:

The CO2 requirement for the plant is:

Amount of CO2 per year = (2.4 lb / KWh)(300,000 KWh)

Amount of CO2 per year = (720000 lb)(1 ton/ 2204.62 lb)

Amount of CO2 per year = 326.59 ton

The perceived economic impact of CO2 generated per year will then be:

Economic Impact = ($25 / ton)(326.59 ton)

<u>Economic Impact = $8164.67</u>

7 0

3 years ago

The design specifications of a 1.2-m long solid circular transmission shaft require that the angle of twist of the shaft not exc

Verizon [17]

Answer:

c = 18.0569 mm

Explanation:

Strategy

We will find required diameter based on angle of twist and based on shearing stress. The larger value will govern.

Given Data

Applied Torque

T = 750 N.m

Length of shaft

L = 1.2 m

Modulus of Rigidity

G = 77.2 GPa

Allowable Stress

г = 90 MPa

Maximum Angle of twist

∅=4°

∅=4* $\pi$ /180

∅=69.813 *10^-3 rad

Required Diameter based on angle of twist

∅=TL/GJ

∅=TL/G* $\pi$ /2*c^4

∅=2TL/G* $\pi$ *c^4

c= $\sqrt[4]{2TL/\pi G }$ ∅

c=18.0869 *10^-3 rad

Required Diameter based on shearing stress

г = T/J*c

г = [T/(J* $\pi$ /2*c^4)]*c

г =[2T/(J* $\pi$ *c^4)]*c

c=17.441*10^-3 rad

Minimum Radius Required

We will use larger of the two values

c= 18.0569 x 10^-3 m

c = 18.0569 mm

3 0

4 years ago