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lesantik [10]
3 years ago
9

Liquid water at 325 K and 8000 kPa flows into a boiler at a rate of 10 kg⋅s− 1 and is vaporized, producing saturated vapor at 80

00 kPa. What is the maximum fraction of the heat added to the water in the boiler that can be converted into work in a process whose product is water at the initial conditions, if Tσ = 300k? what happens to the rest of the heat?
What is the rate of entropy change in the surroundings as a resuit ot the work- producing process? in the system?Total?
Chemistry
1 answer:
masha68 [24]3 years ago
3 0

Answer:

0.4058

Explanation:

From the steam table

At 325 K, saturated liquid

Enthalpy, Hf = 217 kJ/kg

Entropy Sf = 0.7274 kJ/kg-K

Specific volume Vf = 1.013 cm3/gm

Saturated pressure Psat = 12.87 kPa

For compressed liquid

P1 = 8000 kPa

Temperature T = 325 K

Thermal expansion coefficient B = 460 x 10^{-6} K^{-1}

Enthalpy at initial conditions

H1 = Hf + Vf (1-BT)(P1 - P_sat)

= 217 + 1.013*10^-3 (1 - 460*10^{-6})(8000 - 12.87)

= 223.881 kJ/kg

Entropy at initial conditions

S1 = Sf - BVf (P1 - Psat)

= 0.7274 - 460×10^{-6}*1.013*10^-3 (8000 - 12.87)

= 0.724 kJ/kg-K

At 8000 kPa, saturated vapor

H2 = 2759.9 kJ/kg

S2 = 5.7471 kJ/kg-K

T = 300 K

Heat added Q = H_2 - H_1

= 2759.9 - 223.881

= 2536 kJ/kg

Maximum work

W = (H1 - H2) - T (S1 - S2)

= 223.881 - 2759.9 - 300(0.724 - 5.7471)

= - 1029 kJ/kg

Fraction of heat added = W/Q

= 1029/2536

= 0.4058

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A nuclear reactor core must stay at or below 95 °C to remain in good working condition. Cool water at a temperature of 10 °C is
aliina [53]

Answer:

\large \boxed{\text{67 000 g}}

Explanation:

This is a problem in calorimetry — the measurement of the quantities of heat that flow from one object to another.

It is based on the Law of Conservation of Energy — Energy can be transformed from one type to another, but it cannot be destroyed or created.

If heat flows out of the reactor (negative), the same amount of heat must flow into the water (positive).

Since there is no change in total energy,

heat₁ + heat₂ = 0

The symbol for the quantity of heat transferred is q, so we can rewrite the word equation as

q₁ + q₂  = 0

The formula for the heat absorbed or released by an object is

 q = mCΔT, where

 m = the mass of the sample

  C = the specific heat capacity of the sample, and

ΔT = T_f - T_i = the change in temperature

1. Equation

There are two heat flows in this problem,

heat released by reactor + heat absorbed by water = 0

               q₁                  +                        q₂                     = 0

               q₁                  +                 m₂C₂ΔT₂                 = 0

2. Data:

q₁ = -23 746 kJ

m₂ = ?; C₂ = 4.184 J°C⁻¹g⁻¹;  T_f = 95 °C; T_i = 10 °C

3. Calculations

(a) Convert kilojoules to joules

q_{1} = -\text{23 746 kJ} \times \dfrac{\text{1000 J}}{\text{1 kJ}} = -\text{23 746 000 J}

(b) ΔT  

ΔT₂ = T_f - T_i = 95 °C - 10 °C = 85 °C

(c) m₂

\begin{array}{rcl}q_{1} + q_{2} & = & 0\\\text{-23 746 000 J} + m_{2} \times 4.184 \text{ J$^{\circ}$C$^{-1}$g$^{-1}$} \times 85 \, ^{\circ}\text{C} & = & 0\\\text{-23 746 000 J} + 356m_{2} \text{J$\cdot$g}^{-1} & = & 0\\356m_{2} \text{g}^{-1} & = & 23746000\\m_2&=& \dfrac{23746000}{\text{356 g}^{-1}}\\\\ & = & \textbf{67000 g}\\\end{array}\\

\text{You must circulate $\large \boxed{\textbf{67 000 g}}$ of water each hour.}

7 0
3 years ago
What happens to a substance when it dissolves​
uranmaximum [27]

Answer:

A solution is made when one substance called the solute "dissolves" into another substance called the solvent.

Explanation:

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3 years ago
A solution is made by adding 35.5 mL of concentrated hydrochloric acid ( 37.3 wt% , density 1.19 g/mL1.19 g/mL ) to some water i
erastova [34]

Answer:

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Explanation:

We must first obtain the concentration of the concentrated acid from the formula;

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d= density of concentrated acid = 1.19 g/ml

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Substituting values;

Co= 10 × 37.3 × 1.19/36.5

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Co= 12.16 M

We can now use the dilution formula

CoVo= CdVd

Where;

Co= concentration of concentrated acid= 12.16 M

Vo= volume of concentrated acid = 35.5 ml

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Vd= volume of dilute acid = 250ml

Substituting values and making Cd the subject of the formula;

Cd= CoVo/Vd

Cd= 12.16 × 35.5/250

Cd= 1.73 M

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