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1 year ago

There is a picture linked for the question I have if anyone can help me with my whole quiz I’ll try to give 60 points

Engineering

1 answer:

Semmy [17]1 year ago

4 0

Answer:

Where is the picture? I'll try to help as much as I can but I can't promise anything.

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The electrical panel schedules are located on EWR Plan number ___.

Stells [14]

A8 is the answer because yea and because I am a teacher

5 0

2 years ago

A heat pump receives heat from a lake that has an average wintertime temperature of 6o C and supplies heat into a house having a

Dafna1 [17]

Answer:

a) $\dot W = 1.062\,kW$

Explanation:

a) Let consider that heat pump is reversible, so that the Coefficient of Performance is:

$COP_{HP} = \frac{T_{H}}{T_{H}-T_{L}}$

$COP_{HP} = \frac{298.15\,K}{298.15\,K-279.15\,K}$

$COP_{HP} = 15.692$

The minimum heat received by the house must be equal to the heat lost to keep the average temperature constant. Hence:

$\dot Q_{H} = 60000\,\frac{kJ}{h}$

The minimum power supplied to the heat pump is:

$\dot W = \frac{\dot Q_{H}}{COP}$

$\dot W = \frac{\left(60000\,\frac{kJ}{h} \right)\cdot \left(\frac{1\,h}{3600\,s} \right)}{15.692}$

$\dot W = 1.062\,kW$

5 0

2 years ago

A 13.7g sample of a compound exerts a pressure of 2.01atm in a 0.750L flask at 399K. What is the molar mass of the compound?a. 3

Katarina [22]

Answer: Option D) 298 g/mol is the correct answer

Explanation:

Given that;

Mass of sample m = 13.7 g

pressure P = 2.01 atm

Volume V = 0.750 L

Temperature T = 399 K

Now taking a look at the ideal gas equation

PV = nRT

we solve for n

n = PV/RT

now we substitute

n = (2.01 atm x 0.750 L) / (0.0821 L-atm/mol-K x 399 K )

= 1.5075 / 32.7579

= 0.04601 mol

we know that

molar mass of the compound = mass / moles

Molar Mass = 13.7 g / 0.04601 mol

= 297.7 g/mol ≈ 298 g/mol

Therefore Option D) 298 g/mol is the correct answer

4 0

2 years ago

Is santa real or nah is santa real or nah

Elena L [17]

Answer:

nah

Explanation:

3 0

2 years ago

Read 2 more answers

Calculate the molar heat capacity of a monatomic non-metallic solid at 500K which is characterized by an Einstein temperature of

aleksandr82 [10.1K]

Answer:

Explanation:

Given

Temperature of solid $T=500\ K$

Einstein Temperature $T_E=300\ K$

Heat Capacity in the Einstein model is given by

$C_v=3R\left [ \frac{T_E}{T}\right ]^2\frac{e^{\frac{T_E}{T}}}{\left ( e^{\frac{T_E}{T}}-1\right )^2}$

$e^{\frac{3}{5}}=1.822$

Substitute the values

$C_v=3R\times (\frac{300}{500})^2\times (\frac{1.822}{(1.822-1)^2})$

$C_v=3R\times \frac{9}{25}\times \frac{1.822}{(0.822)^2}$

$C_v=0.97\times (3R)$

6 0

2 years ago