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.
Answer:b
Explanation:
Energy is released to form the product
Answer:
[OH⁻] = 2,6x10⁻¹¹
Acidic
Explanation:
The kw in water is:
2 H₂O(l) ⇄ OH⁻(aq) + H₃O⁺(aq)
kw = [OH⁻] [H₃O⁺] = 1,00x10⁻¹⁴
If concentracion of H₃O⁺ is 3,9x10⁻⁴M:
[OH⁻] [3,9x10⁻⁴M] = 1,00x10⁻¹⁴
<em>[OH⁻] = 2,6x10⁻¹¹</em>
pH is defined as - log[H₃O⁺]. If pH>7,0 the solution is basic, if pH<7,0 solution is acidic, if pH=7,0 solution is neutral.
In this problem,
pH = - log [3,9x10⁻⁴M] = <em>3,4</em>
As pH is < 7.0, the solution is <em>acidic</em>
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I hope it helps!
Each element is diffrent than another element
Ok to answer this question we firsst need to fin the number of mol of Urea (CH4N2O). to do this we simply :
1 mol of urea =15/60.055 = 0.25mol
therefore 200g of water contain 0.25mol
the next step is to determine the malality of our solution in 200g of water, to do this we say:
200 g = 1Kg/1000g = 0.2kg
therefor 0.25mol/0.2Kg = 1.25mol/kg
and from the equation:
we know that i = 1
we are given Kf
b is the molality that we just calculated
therefore;
the solutions freezing point is -2.325°C
