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nikdorinn [45]
3 years ago
10

By increasing the length of heat exchanger, the effectiveness and cold stream output temperature of the heat exchanger

Chemistry
1 answer:
Dominik [7]3 years ago
5 0

Answer:

Effectiveness and cold stream output temperature of the heat exchange Increases. So, Answer is b) Increases.

Explanation:

We have a heat exchanger, and it is required to compare the effectiveness and cold stream output if the length is increased.

Heat exchangers are engineering devices used to transfer energy. Thermal energy is transferred from Fluid 1 - Hot fluid (HF) to a Fluid 2 - Cold Fluid (CF). Both fluids 1 and 2 can flow with different values of mass flow rate and different specific heat. When the streams go inside the heat exchanger Temperature of Fluid 1 (HF) will decrease, at the same time Temperature of the Fluid 2 (CF) will increase.

In this case, we need to analyze the behavior taking into account different lengths of heat exchangers. If the length of the heat exchanger increases, it means the transfer area will increases. Heat transfer will increase if the transfer area increases. In this sense, the increasing length is the same than increase heat transfer.

If the heat transfer increases, it means Fluid 1 (HF) will reduce its temperature, and at the same time Fluid 2 (CF) will increase its temperature.

Finally, Answer is b) Effectiveness and cold stream output temperature increases when the length of the heat exchanger is increased.

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Which of the following statements related to S N1 reactions is not true? - The heterolysis of a bond between atoms which do not
Varvara68 [4.7K]

Answer:

The charged carbon atom of a carbocation has a complete octet of valence shell electrons

Explanation:

A charged carbon atom of a carbocation has a valence shell that is not filled, <u>that's why it acts as an electrophile (or a Lewis base)</u>. This unfilled valence shell is also the reason of the nucleophilic attack that takes place during the second step of a SN1 reaction.

6 0
3 years ago
Thermal energy is __________.
Gekata [30.6K]

Answer:

Thermal energy (also called heat energy) is produced when a rise in temperature causes atoms and molecules to move faster and collide with each other. The energy that comes from the temperature of the heated substance is called thermal energy.

Credit:

https://www.solarschools.net/knowledge-bank/energy/types/thermal

5 0
3 years ago
Read 2 more answers
Consider the reaction 2Al + 6HBr → 2AlBr3 + 3H2. If 8 moles of Al react with 8 moles of HBr, what is the limiting reactant?
TiliK225 [7]

Answer:- HBr is limiting reactant.

Solution:- The given balanced equation is:

2Al+6HBr\rightarrow 2AlBr_3+3H_2

From this equation, There is 2:6 mol or 1:3 mol ratio between Al and HBr. Since we have 8 moles of each, HBr is the limiting reactant as we need 3 moles of HBr for each mol of Al.

The calculations could be shown as:

8molAl(\frac{6molHBr}{2molAl})

= 24 mol HBr

From calculations, 24 moles of HBr are required to react completely with 8 moles of Al but only 8 moles of it are available. It clearly indicates, HBr is limiting reactant.

7 0
3 years ago
When 0.5141 g of biphenyl (C12H10) undergoes combustion in a bomb calorimeter, the temperature rises from 25.823 °C to 29.419 °C
natka813 [3]

Answer:

\Delta_{r}U of the reaction is -6313 kJ/mol

\Delta_{r}H of the reaction is -6312 kJ/mol

Explanation:

Temperature\,\,change= \Delta U = 29.419-25.823 =3.506^{o}C

q_{cal}= C \times \Delta T

=5.861 \times 3.596 = 21.076\,kJ

q_{rxn}= -q_{cal}= -21.076\,kJ

\Delta_{r}U= -21.076 \times \frac{154}{0.5141}= -6313\, kJ/mol

Therefore, \Delta_{r}U of the reaction is -6313 kJ/mol.

The chemical reaction in bomb calorimeter  is as follows.

C_{12}H_{10}(s)+\frac{27}{2}O_{2}(g)\rightarrow 12CO_{2}(g)+5H_{2}O(g)

Number\,of\,moles\Delta n=(12+5)-\frac{27}{2}=3.5

\Delta_{r}H=\Delta E+ \Delta n. RT

=-6313+3.5\times 8.314\times 10^{-3} \times 3.596=-6312\,kJ/mol

Therefore, \Delta_{r}H of the reaction is -6312 kJ/mol.

3 0
3 years ago
Determine whether the statement about identifying a halide is true: Regardless of any concentration of ammonium solution, the pr
frosja888 [35]

Answer:

False

Explanation:

The statement ; Regardless of any concentration of ammonium solution the precipitate of unknown halide after 0.1M AgNO3 will remain is FALSE

This is Because the remaining concentration of AgNO3 is dependent on the solubility of Ag⁺

3 0
3 years ago
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