Question 2/5

The air conditioner in a house or a car has a cooler that brings atmospheric air from 30C to 10C, with both states at 101KPa. If

torisob [31]

The rate of heat transfer by the air conditioner using constant specific heat of 1.004kj/kg.K is 15.06 kW.

<h3>What is the rate of heat transfer?</h3>

Rate of heat transfer is the power rating of the machine.

Work done and changes in potential and kinetic energy are neglected since it is a steady state process.

The specific heat in terms of specific heat capacity and temperature change is given as:

$q_{out} = Cp(Ti - Te)$

$q_{out} = 1.004(30 - 10) = 20.08 kJ/kg \\$

The rate of heat transfer, is then determined as follows:

Qout = flow rate × specific heat

Qout = 0.75 × 20.08 = 15.06 kW

Therefore, the rate of heat transfer by the air conditioner is 15.06 kW.

Learn more about rate of heat transfer at: brainly.com/question/17152804

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6 0

2 years ago

A ________ can be installed in a cast-iron block to repair a worn or cracked cylinder. Question 24 options:

VLD [36.1K]

Answer:

Cylinder sleeve

Explanation:

7 0

2 years ago

A spring-loaded piston-cylinder contains 1 kg of carbon dioxide. This system is heated from 104 kPa and 25 °C to 1,068 kPa and 3

labwork [276]

Answer:

Q = -68.859 kJ

Explanation:

given details

mass $co_2 = 1 kg$

initial pressure P_1 = 104 kPa

Temperature T_1 = 25 Degree C = 25+ 273 K = 298 K

final pressure P_2 = 1068 kPa

Temperature T_2 = 311 Degree C = 311+ 273 K = 584 K

we know that

molecular mass of $co_2 = 44$

R = 8.314/44 = 0.189 kJ/kg K

c_v = 0.657 kJ/kgK

from ideal gas equation

PV =mRT

$V_1 = \frac{m RT_1}{P_1}$

$=\frac{1*0.189*298}{104}$

$V_1 = 0.5415 m3$

$V_2 = \frac{m RT_2}{P_2}$

$=\frac{1*0.189*584}{1068}$

$V_1 = 0.1033 m3$

WORK DONE

$W =P_{avg}*{V_2-V_1}$

w = 586*(0.1033 -0.514)

W =256.76 kJ

INTERNAL ENERGY IS

$\Delta U = m *c_v*{V_2-V_1}$

$\Delta U = 1*0.657*(584-298)$

$\Delta U =187.902 kJ$

HEAT TRANSFER

$Q = \Delta U +W$

= 187.902 +(-256.46)

Q = -68.859 kJ

7 0

3 years ago

A steel bar 110 mm long and having a square cross section 22 mm on an edge is pulled in tension with a load of 89,000 N, and exp

iragen [17]

Answer:

Elastic modulus of steel = 202.27 GPa

Explanation:

given data

long = 110 mm = 0.11 m

cross section 22 mm = 0.022 m

load = 89,000 N

elongation = 0.10 mm = 1 × $10^{-4}$ m

solution

we know that Elastic modulus is express as

Elastic modulus = $\frac{stress}{strain}$ ................1

here stress is

Stress = $\frac{Force}{area}$ .................2

Area = (0.022)²

and

Strain = $\frac{extension}{length}$ .............3

so here put value in equation 1 we get

Elastic modulus = $\frac{89,000\times 0.11}{0.022^2 \times 1 \times 10^{-4} }$

Elastic modulus of steel = 202.27 × $10^9$ Pa

Elastic modulus of steel = 202.27 GPa

3 0

3 years ago

If a vacuum gau ge reads 9.62 psi, it means that: a. the very highest column of mercury it could support would be 19.58 inches.

scZoUnD [109]

Answer:All of the above

Explanation:

9.62 psi means 497.49 mm of Hg pressure

for (a)19.58 inches is equals to 497.49 mm of Hg

(b)atmospheric pressure is 14.69 psi

vaccum gauge is 9.62psi

absolute pressure is=14.69-9.62=5.07

(c)vaccum means air is sucked and there is negative pressure so it tells about below atmospheric pressure.

thus all are correct

8 0

3 years ago