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galina1969 [7]
3 years ago
9

What happens to the outlet temperatures of the hot and cold fluid when the both the hot fluid is at a maximum and the cold fluid

is at a minimum flowrate?
a- Remains same as that of inlet temperature

b- Hot water temperature decreases and cold water temperature increase

c- Hot water remains almost same but cold water temperature increases drastically.
Chemistry
1 answer:
Alina [70]3 years ago
7 0

Answer:

c- Hot water remains almost same but cold water temperature increases drastically.

Explanation:

In a process of heat exchange where a hot and a cold fluid are present, the amount of heat that is transfered from one to another, the heat lost by the hot fluid, is gained by the cold fluid. That can be calculated by the equation : Q = m * Cp * dT.

Where:

m= mass flowrate

Cp=specific-heat-capacity

dT= difference between the inlet and outlet temperatures

We can state the equality of this equation for both fluids:

(mass-flowrate * specific-heat-capacity * (temperature-in – temperature-out) )<em> for hot medium</em> = (mass-flowrate * specific-heat-capacity * (temperature-out – temperature-in) )<em> for cold medium</em>

Now if we increase the value of the mass flow rate of the hot fluid to its maximum and decrease the mass flowrate of the cold fluid to its minimum. That will mean that the difference between the inlet and outlet temperatures for the cold fluid must increase so as to compensate the increase on the other side of the equation and guarantee the equality. And taking in account that the inlet temperature of the cold fluid is known and what can change is the outlet one, the correct answer is c that says that the outlet temperature of the cold fluid increases drasically.

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Which solution has the lowest freezing point?
Leokris [45]
<span>Correct Answer: Option 3 i.e. 30 g of KI dissolved in 100 g of water.

Reason:
Depression in freezing point is a colligative property and it is directly proportional to molality of solution.

Molality of solution is mathematically expressed as,
Molality = </span>\frac{\text{Number of moles}}{\text{Weight of solvent (Kg)}}<span>

In case of option 1 and 2, molality of solution is 0.602 m. For option 3, molality of solution is 1.807 m, while in case of option 4, molality of solution is 1.205 m.

<u><em>Thus, second solution (option 2) has highest concentration (in terms of molality). Hence, it will have lowest freezing point</em></u></span>
3 0
4 years ago
Read 2 more answers
Since a solution of KBr would probably cause significant damage to the car radiator and engine, you decide to use 125 g of the n
Vilka [71]

Answer:

When octane is used, the solution will have less effect on the freezing point depression of the solution

Explanation:

The complete question is:

Calculate the freezing point of a solution of 125 g KBr in 450 g water.

Since a solution of KBr would probably cause significant damage to the car radiator and engine, you decide to use 125 g of the nonelectrolyte octane (molar mass 114 g/mole). Will this have a greater or less effect on freezing point depression of the solution?

Step 1: Data given

Molar mass of KBr = 119.0 g/mol

Molar mass of octane = 114 g/mol

Mass of KBr = 125 grams

Mass of octane = 125 grams

Mass of water = 450 grams

Step 2: Calculate moles KBr

Moles KBr = mass KBr / molar mass KBr

Moles KBr = 125 grams / 119.0 g/mol

Moles KBr = 1.05 moles

Step 3: Calculate moles octane

Moles octane = 125 grams / 114 g/mol

Moles octane = 1.10 moles

Step 4: Calculate molality

Molality = moles compound / mass water

Molality KBr = 1.05 moles / 0.450 kg

Molality KBr = 2.33 molal

Molality octane = 1.10 moles / 0.450 kg

Molality octane = 2.44  molal

Step 5: Calculate the freezing point depression when KBr is used

ΔT = i*Kf * m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of KBr = 2

⇒with Kf = the freezing point depression constant of water = 1.86 °C/m

⇒with m= the molality = 2.33 molal

ΔT = 2*1.86 * 2.33

ΔT = 8.68 °C

This means the freezing point of this solution is -8.68 °C

Step 6: Calculate the freezing point depression when octane is used

ΔT = i*Kf * m

⇒with ΔT = the freezing point depression = TO BE DETERMINED

⇒with i = the van't Hoff factor of the nonelectrolyte octane = 1

⇒with Kf = the freezing point depression constant of water = 1.86 °C/m

⇒with m= the molality = 2.44 molal

ΔT = 1* 1.86 * 2.44

ΔT = 4.54 °C

This means the freezing point of this solutions is -4.54 °C

When octane is used, the solution will have less effect on the freezing point depression of the solution

7 0
3 years ago
a weather balloon contains 8.80 moles of helium at a ppressure of 0.992 atm and a temperature os 25 C at ground level. What is t
djverab [1.8K]
We can use the ideal gas law equation to find the volume of the balloon.
PV = nRT 
where 
P - pressure - 0.992 atm x 101 325 Pa/atm = 100 514 Pa
V - volume 
n - number of moles - 8.80 mol 
R - universal gas constant  - 8.314 Jmol⁻¹K⁻¹
T - temperature in kelvin - 25 °C + 273 = 298 K
Substituting these values in the equation 
100 514 Pa x V = 8.80 mol x  8.314 Jmol⁻¹K⁻¹ x 298 K
V = 217 L
volume of balloon is 217 L
7 0
3 years ago
Please answer this it’s due tomorrow I’ll give you brainliest
Ipatiy [6.2K]

Answer:

The answer is the last one

Explanation:

1. Give

2.me

3. Brainliest

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3 0
3 years ago
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hurry please! a mixture of He, Ne, and N2 gases has a pressure of 1.348 atm. if the pressures of He and Ne are 0.124 atm and 0.9
Murrr4er [49]

Answer:

A. 0.228

Explanation:

Partial pressure as name suggests is partial pressures that add up to a final (total) pressure.

Here total pressure is 1.348 atm.

He + Ne + N2 = 1.348

0. 124 + 0.996 + N2 = 1.348

N2 = 0.228

3 0
4 years ago
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