Answer:
=> 572.83 K (299.83°C).
=> 95.86 m^2.
Explanation:
Parameters given are; Water flowing= 13.85 kg/s, temperature of water entering = 54.5°C and the temperature of water going out = 87.8°C, gas flow rate 54,430 kg/h(15.11 kg/s). Temperature of gas coming in = 427°C = 700K, specific heat capacity of hot gas and water = 1.005 kJ/ kg.K and 4.187 KJ/kg. K, overall heat transfer coefficient = Uo = 69.1 W/m^2.K.
Hence;
Mass of hot gas × specific heat capacity of hot gas × change in temperature = mass of water × specific heat capacity of water × change in temperature.
15.11 × 1.005(700K - x ) = 13.85 × 4.187(33.3).
If we solve for x, we will get the value of x to be;
x = 572.83 K (2.99.83°C).
x is the temperature of the exit gas that is 572.83 K(299.83°C).
(b). ∆T = 339.2 - 245.33/ln (339.2/245.33).
∆T = 93.87/ln 1.38.
∆T = 291.521K.
Heat transfer rate= 15.11 × 1.005 × 10^3 (700 - 572.83) = 1931146.394.
heat-transfer area = 1931146.394/69.1 × 291.521.
heat-transfer area= 95.86 m^2.
Answer:
9.1
Explanation:
Step 1: Calculate the basic dissociation constant of propionate ion (Kb)
Sodium propionate is a strong electrolyte that dissociates according to the following equation.
NaC₃H₅O₂ ⇒ Na⁺ + C₃H₅O₂⁻
Propionate is the conjugate base of propionic acid according to the following equation.
C₃H₅O₂⁻ + H₂O ⇄ HC₃H₅O₂ + OH⁻
We can calculate Kb for propionate using the following expression.
Ka × Kb = Kw
Kb = Kw/Ka = 1.0 × 10⁻¹⁴/1.3 × 10⁻⁵ = 7.7 × 10⁻¹⁰
Step 2: Calculate the concentration of OH⁻
The concentration of the base (Cb) is 0.24 M. We can calculate [OH⁻] using the following expression.
[OH⁻] = √(Kb × Cb) = √(7.7 × 10⁻¹⁰ × 0.24) = 1.4 × 10⁻⁵ M
Step 3: Calculate the concentration of H⁺
We will use the following expression.
Kw = [H⁺] × [OH⁻]
[H⁺] = Kw/[OH⁻] = 1.0 × 10⁻¹⁴/1.4 × 10⁻⁵ = 7.1 × 10⁻¹⁰ M
Step 4: Calculate the pH of the solution
We will use the definition of pH.
pH = -log [H⁺] = -log 7.1 × 10⁻¹⁰ = 9.1
Answer:- 6984 kJ of heat is produced.
Solution:- From given information, 1367 kJ of heat is produced by the combustion of 1 mole of ethanol. We are asked to calculate the heat produced by the combustion of 235.0 g of ethanol.
Let's convert given grams to moles and multiply by the heat produced by one mole of ethanol to get the total heat produced. Molar mass of ethanol is 46 grams per mole. The set will be:
= 6984 kJ
So, 6984 kJ of heat is produced by the combustion of 235.0 g of liquid ethanol.
4
Explanation:
The number 4 in the hydrologic cycle depicts surface runoff.
The hydrologic cycle depicts the cyclic movement of water between different earth components.
It shows water in all its phases.
- From the attached image, runoff is depicted 4.
- Runoff is water that moves on the surface after precipitation.
- Number 1 is evaporation where water changes to vapor.
- Number 2 is condensation where the vapor cools back to liquid.
- Number 3 is precipitation where the liquid water is forced down under gravity.
- On the surface, the water runs as runoff.
learn more:
Biogeochemical cycle brainly.com/question/3509510
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