What kind of material will reduce the flow of heat energy?

what is the reason for the difference between the values of the dry adiabatic lapse rate and the moist adiabatic lapse rate?

aalyn [17]

Latent heat released during condensation offsets some of the cooling of a rising air parcel, causing the moist adiabatic lapse rate to have a smaller value.

6 0

2 years ago

Why is there no overall electrical charge on an aluminium atom?

Alex787 [66]

The atoms always have an equal amount of protons and neutrons so lets say if one had 10 positive then it would have 10 negatives making it neutral.
10-10=0

3 0

3 years ago

If 13.7 grams of manganese oxide reacts with excess hydrogen chloride gas, how many grams of water are formed?

Lerok [7]

Answer:

5.67 g OF WATER WILL BE FORMED WHEN 13.7 g OF MnO2 REACTS WITH HCl GAS.

Explanation:

EQUATION FOR THE REACTION

Mn02 + 4HCl --------> MnCl2 + Cl2 + 2H2O

From the balanced reaction between manganese oxide and hydrogen chloride gas;

1 mole of MnO2 reacts to form 2 mole of water

At STP, the molecular mass of the sample is equal to the mole of the substance. So therefore:

(55 + 16 * 2) g of MnO2 reacts to form 2 * ( 1 *2 + 16) g of water

(55 + 32) g of MnO2 reacts to form 2 * 18 g of water

87 g of MnO2 reacts to form 36 g of water

If 13.7 g of MnO2 were to be used?

87 g of MnO2 = 36 g of H2O

13.7 g of MnO2 = ( 13.7 * 36 / 87) g of water

= 493.2 / 87 g of water

Mass of water = 5.669 g of water

Approximately 5.67 g of water will be formed when 13.7 g of manganese oxide reacts with excess hydrogen chloride gas.

3 0

3 years ago

If 8.500 g CH is burned and the heat produced from the burning is added to 5691 g of water at 21 °C, what is the final

mojhsa [17]

The final temperature = 36 °C

<h3>Further explanation</h3>

The balanced combustion reaction for C₆H₆

2C₆H₆(l)+15O₂(g)⇒ 12CO₂(g)+6H₂O(l) +6542 kJ

MW C₆H₆ : 78.11 g/mol

mol C₆H₆ :

$\tt \dfrac{8.5}{78.11}=0.109$

Heat released for 2 mol C₆H₆ =6542 kJ, so for 1 mol

$\tt \dfrac{0.109}{2}\times 6542=356.539~kJ/mol$

Heat transferred to water :

Q=m.c.ΔT

$\tt 356.539=5.691~kg\times 4.18~kj/kg^oC\times (t_2-21)\\\\t_2-21=15\rightarrow t_2=36^oC$

3 0

2 years ago

An ideal gas (C}R), flowing at 4 kmol/h, expands isothermally at 475 Kfrom 100 to 50 kPa through a rigid device. If the power pr

Zina [86]

<u>Answer:</u> The rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

<u>Explanation:</u>

We are given:

$C_p=\frac{7}{2}R\\\\T=475K\\P_1=100kPa\\P_2=50kPa$

Rate of flow of ideal gas , n = 4 kmol/hr = $\frac{4\times 1000mol}{3600s}=1.11mol/s$ (Conversion factors used: 1 kmol = 1000 mol; 1 hr = 3600 s)

Power produced = 2000 W = 2 kW (Conversion factor: 1 kW = 1000 W)

We know that:

$\Delta U=0$ (For isothermal process)

So, by applying first law of thermodynamics:

$\Delta U=\Delta q-\Delta W$

$\Delta q=\Delta W$ .......(1)

Now, calculating the work done for isothermal process, we use the equation:

$\Delta W=nRT\ln (\frac{P_1}{P_2})$

where,

$\Delta W$ = change in work done

n = number of moles = 1.11 mol/s

R = Gas constant = 8.314 J/mol.K

T = temperature = 475 K

$P_1$ = initial pressure = 100 kPa

$P_2$ = final pressure = 50 kPa

Putting values in above equation, we get:

$\Delta W=1.11mol/s\times 8.314J\times 475K\times \ln (\frac{100}{50})\\\\\Delta W=3038.45J/s=3.038kJ/s=3.038kW$

Calculating the heat flow, we use equation 1, we get:

[ex]\Delta q=3.038kW[/tex]

Now, calculating the rate of lost work, we use the equation:

$\text{Rate of lost work}=\Delta W-\text{Power produced}\\\\\text{Rate of lost work}=(3.038-2)kW\\\text{Rate of lost work}=1.038kW$

Hence, the rate of heat flow is 3.038 kW and the rate of lost work is 1.038 kW.

4 0

3 years ago