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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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Hence, Option (1) is correct answer.

<h3>What is Latent Heat ? </h3>

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Consider the following unbalanced chemical equation. C5H12(l) + O2(g) → CO2(g) + H2O(l) If 21.9 grams of pentane (C5H12) are bur
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Answer : The mass of water produced will be 32.78 grams.

Explanation : Given,

Mass of C_5H_{12} = 21.9 g

Molar mass of C_5H_{12} = 72.15 g/mole

Molar mass of H_2O = 18 g/mole

First we have to calculate the moles of C_5H_{12}.

\text{Moles of }C_5H_{12}=\frac{\text{Mass of }C_5H_{12}}{\text{Molar mass of }C_5H_{12}}=\frac{21.9g}{72.15g/mole}=0.3035moles

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The balanced chemical reaction will be,

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From the balanced reaction we conclude that

As, 1 mole of C_5H_{12} react to give 6 moles of H_2O

So, 0.3035 moles of C_5H_{12} react to give 0.3035\times 6=1.821 moles of H_2O

Now we have to calculate the mass of H_2O.

\text{Mass of }H_2O=\text{Moles of }H_2O\times \text{Molar mass of }H_2O

\text{Mass of }H_2O=(1.821mole)\times (18g/mole)=32.78g

Therefore, the mass of water produced will be 32.78 grams.

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