A 400-m^3 storage tank is being constructed to hold LNG, liquefied natural gas, which may be assumed to be essentially pure methane. If the tank is to contain 90% liquid and 10% vapor, by volume, at 150 K, what mass of LNG (kg) will the tank hold? What is the quality in the tank?

Answer:

Yes

Explanation:

Type of control system

1.Open system

 When mass and energy is allowed to pass through the system ,then this type of system is called open system.

2.Close system

 When only energy is allowed to pass through the system ,then this type of system is called close system.

3.Isolated system

 When mass and energy did not allowed to pass through the system ,then this type of system is called isolated system.

In the engine radiator ,mass of hot fluid enters and interact with the cold air then after heat transfer take place between these two fluids,that is why we can say that radiator is an open system.

It become necessity to use the radiator in all type of engine because it produce cooling effect to the engine cylinder .If temperature of engine cylinder will become too high then it will leads to the high thermal stresses and may be it will break the engine cylinder.

Yes, the radiator should be analyzed as an open system.

In thermodynamics, a system is classified as open if both mass and energy can cross its boundaries. The radiator in a car's engine cooling system is designed to dissipate heat from the engine to the air. This dissipation occurs through the circulation of water (or coolant) which absorbs heat from the engine and releases it to the air flowing through the radiator.

In the radiator:

1. Mass Flow: The coolant continuously enters and exits the radiator, carrying thermal energy with it.

2. Energy Transfer: Heat is transferred from the hot coolant to the cooler ambient air through the radiator fins, driven by forced convection due to airflow.

3. Open System Characteristics: Both mass (coolant) and energy (heat) cross the boundaries of the radiator.

Considering these aspects, the radiator facilitates the flow of coolant (mass) and the transfer of thermal energy (heat), making it an open system.

The radiator in a car should be analyzed as an open system due to the continuous flow of coolant and heat transfer involved in its operation. This perspective is essential for accurately understanding and optimizing the cooling process.

Answer:

[tex]T_4[/tex]= 506.79 K

Explanation:

Given that

Minimum temperature [tex]T_1[/tex]= 25 C

Pressure = 100 KPa

Highest temperature [tex]T_3[/tex]= 1000 C

Compression ratio r= 10

We know that for ideal Otto cycle

[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]

Where  [tex]T_3[/tex] is the exhaust temperature.

Now by putting the values

[tex]\dfrac{T_3}{T_4}=r^{\gamma -1}[/tex]

[tex]\dfrac{1000+273}{T_4}=10^{1.4-1}[/tex]

[tex]T_4[/tex]= 506.79 K

Answer:

P=7.027psi

V=1.9788ft ^3

Explanation:

Through laboratory tests, thermodynamic tables were developed, these allow to know all the thermodynamic properties of a substance (entropy, enthalpy, pressure, specific volume, internal energy etc ..)

through prior knowledge of two other properties such as pressure and temperature.

For the first part of this problem, we must find the water saturation pressure at a temperature of 177F

Psat @(177F)=7.027psi

For the second part of this problem, we calculate the specific volume of the water knowing the temperature and that the state is saturated liquid, then multiply by the mass to know the volume

v@177F=0.01649ft^3/lbm

V=mv

V=(0.01649ft^3/lbm)(120lbm)

V=1.9788ft ^3

Answer:

13282.3 years

Explanation:

The C-14 decays exponentially:

[tex]\frac{dN}{dt} =λ*N[/tex]

The solution for this equation is  

[tex]N= N_{o}*e^{- λt}[/tex]

Where:  

No = atom number of C-14 in t=0

N = atom number of C-14 now  

I= radioactive decay constant  

clearing t this equation we get:  

[tex]t=-\frac{1}{ λ}*ln\frac{N}{N_{o}}[/tex]

The term 1/I is called half-life and the value for C-14 is 8252 years.  

N for this exercise is 0.2No  

[tex] t= -8033 * ln \frac{0.2N_{o} }{N_{o}}[/tex]

t = 13282.3 years

Answer:

Diameter of riser =6.02 mm

Explanation:

Given that

Dimensions of rectangular plate is 200mm x 100mm x 20mm.

Volume of rectangle V= 200 x 100 x 20 [tex]mm^3[/tex]

Surface area of rectangle A

A=2(200 x 100+100 x 20 +20 x 200)[tex]mm^2[/tex]

So V/A=7.69

We know that

Solidification times given as

[tex]t=K\left(\dfrac{V}{A}\right)^2[/tex]   -----1

Lets take diameter of riser is d

Given that riser is in spherical shape so V/A=d/6

And

Time for solidification of rectangle is 3.5 min then time for solidificartion of riser is 4.2 min.

Lets take [tex]\dfrac{V}{A}=M[/tex]

[tex]\dfrac{M_{rac}}{M_{riser}}=\dfrac{7.69}{\dfrac{d}{6}}[/tex]

Now from equation 1

[tex]\dfrac{3.5}{4.2}=\left(\dfrac{7.69}{\dfrac{d}{6}}\right)^2[/tex]

So by solving this d=6.02 mm

So the diameter of riser is 6.02 mm.

Answer:

The pressure in pounds per square inch (psi) equals 14.50 psi.

Explanation:

Since psi stands for 'pounds per square inch' and Pascals stands for 'Newtons per square meters'

We can write the given pressure as

[tex]Pressure=100kPa\\\\Pressure=100\times 10^{3}N/m^{2}[/tex]

Now since 1 Newton of force equals 0.22481 Pound force also

since 1 meter equals 39.37 inches

Using the above values we get

[tex]Pressure=100\times 10^{3}\times \frac{0.22481lb}{(39.37)^{2}inch^{2}}[/tex]

thus the pressure becomes

[tex]Pressure=100\times 10^{3}\times \frac{0.22481}{(39.31)^{2}}pounds/inch^{2}\\\\Pressure=14.50psi[/tex]

Answer:

minimum flow rate provided by pump is 0.02513 m^3/s

Explanation:

Given data:

Exit velocity of nozzle = 20m/s

Exit diameter = 40 mm

We know that flow rate Q is given as

[tex]Q = A \times V[/tex]

where A is Area

[tex]A =\frac{\pi}{4} \times (40\times 10^{-3})^2 = 1.256\times 10^{-3} m^2[/tex]

[tex]Q = 1.256\times 10^{-3} \times 20 = 0.02513 m^3/s[/tex]

minimum flow rate provided by pump is 0.02513 m^3/s

Answer:

[tex]1\frac{1}{2}[/tex] in dia rod

Explanation:

Given;

Maximum tensile stress = 20,000 psi

Applied tensile load = 35,000 lb

Now,

the area of the steel rod required = [tex]\frac{\textup{Applied tensile load}}{\textup{Maximum tensile stress}}[/tex]

or

the area of the steel rod required = [tex]\frac{\textup{35,000}}{\textup{20,000}}[/tex]

or

the area of the steel rod required = 1.75 in²

Also, Area of rod = [tex]\frac{\pi}{4}d^2[/tex]

where, d is the diameter of the rod

1.75 in² = [tex]\frac{\pi}{4}d^2[/tex]

or

d = 1.49 in

So we can use [tex]1\frac{1}{2}[/tex] in dia rod

Answer:

Explanation:

In aeronautics stall is the separation of the boundary layer from the wings of plane. This causes loss of lift because the otherwise low pressure area above the wing become filled with turbulent whirls.

Stall occurs when the critical angle of attack of the wing is exceeded.

Answer:

1072 acre foot

1331424000 kg

Explanation:

1 feet has 12 inches, so 2 in is 0.167 feet.

1 km^2 has 1 million m^2.

1 acre is 4074 m^2.

So, 1 km is 247 acres.

Then 26 km^2 is 6422 acres.

So, the volume of water is

6422 * 0.167 = 1072 acre-foot

Since one cubic meter of water has 1000 kg

One inch is 25.4 mm = 0.0254 m

One feet is 12 * 0.0254 = 0.3048 m

An acre-feet has a volume of

4074*0.3048 = 1242 m^3

And that is a mass of water of

1242 * 1000 = 1242000 kg/acre-feet

Therefore the mass of rainwater in the town is of

1072 * 1242000 = 1331424000 kg = 1331424 tons

Answer:

2.9*10^14 electrons

Explanation:

An Ampere is the flow of one Coulomb per second, so 35 μA = is 35*10^-6 C per second.

An electron has a charge of 1.6*10^-19 C.

35*10^-6 / 1.6*10^-19 = 2.9*10^14 electrons

So, with a current o 35 μA you have an aevrage of 2.9*10^14 electrons flowing past a fixed reference cross section perpendicular to the direction of flow.

Answer:

[tex]a=0.0937 \ m/s^2[/tex]

Explanation:

Given that

Angular velocity (ω)= 0.6 rad/s

Angular acceleration (α)= 0.3 [tex]rad/s^2[/tex]

Radius (r)= 0.2 m

We know that is disc is rotating and having angular acceleration then it will have two acceleration .one is radial acceleration and other one is tangential acceleration.

So

[tex]Radial\ acceleration(a_r)=\omega ^2r\ m/s^2[/tex]

[tex]Radial\ acceleration(a_r)=0.6^2 \times0.2 \ m/s^2[/tex]

[tex](a_r)=0.072 \ m/s^2[/tex]

[tex]Tangential\ acceleration(a_t)=\alpha r\ m/s^2[/tex]

[tex]Tangential\ acceleration(a_t)=0.3\times 0.2\ m/s^2[/tex]

[tex](a_t)=0.06 \ m/s^2[/tex]

So the total acceleration ,a

[tex]a=\sqrt{a_t^2+a_r^2}\ m/s^2[/tex]

[tex]a=\sqrt{0.06^2+0.072^2}\ m/s^2[/tex]

[tex]a=0.0937 \ m/s^2[/tex]

Answer:

The main component of fixed wing aircraft

1.Wings

2.Fuselages

3.Landing gear

4.Stabilizers

5.Flight control surface

Cantilever wing:

  A cantilever wing is directly attached to the fuselages and do not have any external support.

Semi cantilever wing:

A cantilever wing does not directly attached to the fuselages and  have any external support.The external support may be one two.

The question involves calculating the duration fuel will last in a power station with 12% efficiency given its energy content and the daily energy demand. A misunderstanding in the conversion of units and efficiency calculation was identified in the attempt, highlighting the need for a corrected mathematical approach involving unit conversions and understanding power station efficiency.

A certain amount of fuel contains 15×102 Btu of energy and is converted into electric energy in a power station having a 12 percent overall efficiency. The average daily demand on the station is 5MW. To find out in how many days the fuel will be totally consumed, first, it's essential to calculate the total energy converted into electrical energy considering the efficiency of the power station. Given the power station efficiency, only 12% of the fuel energy is converted into electrical energy. The total energy available for conversion is 15×102 Btu.

First, convert 15×102 Btu into megajoules (MJ) as 1 Btu is approximately equal to 0.00105506 MJ. Therefore, the energy in MJ is 15×102 × 0.00105506 = 1.58259 MJ. Given that 12% of this energy is usable for electricity, the usable energy is 1.58259 MJ × 12% = 0.189911 MJ. Since 1 MW is 1 MJ/sec, and given the station's demand is 5MW over 24 hours, the total consumption is 5MW × 24h = 120 MW/day.

Then, to find out how many days the fuel would last, divide the total usable electrical energy by the daily consumption. However, an error in the calculation process prevents us from completing the result as intended initially. This calculation seems to involve a misunderstanding regarding the conversion of units and the calculation of efficiency. Therefore, a corrected approach should involve accurately converting the initial Btu to MJ or kWh, taking into account the efficiency correctly, and then calculating the duration of fuel consumption based on the daily energy output and demand.

Answer:

0.08kg/s

Explanation:

For this problem you must use 2 equations, the first is the continuity equation that indicates that all the mass flows that enter is equal to those that leave the system, there you have the first equation.

The second equation is obtained using the first law of thermodynamics that indicates that all the energies that enter a system are the same that come out, you must take into account the heat flows, work and mass flows of each state, as well as their enthalpies found with the temperature.

finally you use the two previous equations to make a system and find the mass flows

I attached procedure

Answer:

P=41.84psi

Explanation:

Cavitation is a phenomenon that arises when the pressure of a liquid fluid reaches the vapor pressure, this means that if at any point in a system of pipes and pumps the water pressure is equal to or less than the vapor pressure it begins to evaporate, producing explosive bubbles that can cause damage and noise in the system.

Therefore, for this problem, all we have to do is find the water vapor pressure at a temperature of 270F, using  thermodynamics tables

Pv@270F=41.84psi

Answer:

32 horsepower will be equal to 17600 ft-lb/sec

Explanation:

We have given 1 horsepower = 550 ft-lb/sec

We have to convert 32 horsepower into ft-lb/sec

As we know that 1 horsepower is equal to 550 ft-lb/sec

So for converting horsepower to ft-lb/sec we have multiply with 550

We have given 32 horse power

So [tex]32\ horsepower =32\times 550=17600ft-lb/sec[/tex]

So 32 horsepower will be equal to 17600 ft-lb/sec

Answer:

[tex]p = 15260.643 \ lbf/ft^2[/tex]

Explanation:

person weight is 175 lbm

weight of stake board 4lbm

size of stakeboard = 2ft by 0.8 ft

area of stakeboard is [tex]2*0.8 ft^2 = 1.6 ft^2[/tex]

gauge pressure is given as

[tex]p =\frac{ w_p+w_s}{A} g[/tex]

where is g is acceleration due to gravity =  32.17 ft/sec^2

puttng all value to get pressure value

   [tex]= \frac{175 +4}{1.6}* 32.17[/tex]

[tex]p = 15260.643 \ lbf/ft^2[/tex]

Answer:

a)Ek=115759.26J

b)23.15kW

Explanation:

kinetic energy(Ek) is understood as that energy that has a body with mass when it travels with a certain speed, it is calculated by the following equation

[tex]Ek=0.5mv^{2}[/tex]

for the first part

m=1200Kg

V=50km/h=13.89m/s

solving for Ek

[tex]Ek=0.5(1200)(13.89)^2[/tex]

Ek=115759.26J

For the second part of the problem we calculate the kinetic energy, using the same formula, then in order to find the energy change we find the difference between the two, finally divide over time (15s) to find the power.

m=1200kg

V=100km/h=27.78m/s

[tex]Ek=0.5(1200)(27.78)^2=462963J\\[/tex]

taking into account all of the above the following equation is inferred

ΔE=[tex]\frac{Ek2-EK1}{T}=\frac{462963-115759.26}{15}  =23146.916W=23.15kW[/tex]

Answer:

The concentration in mg/L equals 0.01 mg/L

Explanation:

Since the concentration is given as 10 parts per billion

We know that

1 part per billion equals 0.001 mg/L

Thus 10 parts per billion  concentration equals

[tex]10\times 0.001mg/L\\\\=0.01mg/L[/tex]

Answer:

acceleration is 0.2181 m/s²

acceleration is 27.05 m/s²

Explanation:

given data

s = 0

time t = 0

velocity V = 5 sin (4s²) m/s

to find out

particle's acceleration

solution

we have given velocity = 5 sin(4s²)

and we know here that velocity = [tex]\frac{ds}{dt}[/tex]

so acceleration will be a =   [tex]\frac{dv}{dt}[/tex]

put here velocity v

acceleration =  [tex]\frac{dv}{dt}[/tex]

acceleration =  [tex]\frac{d(5sin(4s^4))}{dt}[/tex]

acceleration = 5 cos4s² × 8s × [tex]\frac{ds}{dt}[/tex]

acceleration =  5 cos4s² × 8s × 5 sin4s²

acceleration = 200 s ×cos4s²  × sin4s²

put here s = 0.25 and s = 1.25

so

acceleration = 200× 0.25 ×cos(4×0.25²)  × sin(4×0.25²)

acceleration = 0.2181 m/s²

acceleration = 200× 1.25 ×cos(4×1.25²)  × sin(4×1.25²)

acceleration = 27.05 m/s²

The energy used by the motor for one year is 102,313,596,240 ft.lb/year, expressed in both ft.lb and W.h.

1 horsepower (hp) is approximately equal to 745.7 watts (W).

So,[tex]12.5 hp * 745.7 W/hp[/tex]

= 9312.5 W.

The motor draws 9312.5 watts for 16 hours per day.

So, [tex]9312.5 W * 16 hours/day[/tex]

= 148,200 W.h/day.

Multiply the daily energy by the number of days per year:

The motor operates for 5 days per week, so in one year (considering 52 weeks), it operates for 5 days/week × 52 weeks/year = 260 days/year.

[tex]148,200 W.h/day *260 days/year[/tex]

= 38,532,000 W.h/year.

To express the result in ft.lb, we need to convert the energy from watt-hours to foot-pounds:

1 watt-hour (Wh) is approximately equal to 2655.22 foot-pounds (ft.lb).

So, [tex]38,532,000 W.h/year * 2655.22 ft.lb/W.h[/tex]

≈ 102,313,596,240 ft.lb/year.

Answer:

work done = 48.88 × [tex]10^{9}[/tex] J

Explanation:

given data

mass = 100 kN

velocity =  310 m/s

time = 30 min = 1800 s

drag force = 12 kN

descends = 2200 m

to find out

work done by the shuttle engine

solution

we know that work done here is

work done = accelerating work - drag work - descending work

put here all value

work done = ( mass ×velocity ×time  - force ×velocity ×time  - mass ×descends )  10³ J

work done = ( 100 × 310 × 1800  - 12×310 ×1800  - 100 × 2200 )  10³ J

work done = 48.88 × [tex]10^{9}[/tex] J

Answer:

a)True

Explanation:

In cutting action,three type of zone are presents

1.Primary zone :

The most of part of energy is converted in to heat.

2.Secondary zone:

Heat generation due to rubbing between tool and chip.This also called deformation zone.

3.Tertiary zone:

Heat generated due to flank of tool and already machined surface.

So the highest temperature is located close to the shear zone.

Answer:

(A) Dimension of force  [tex]MLT^{-2}[/tex]

(b) Dimension of energy [tex]ML^2T^{-2}[/tex]

(c) Dimension of power = [tex]ML^2T^{-3}[/tex]

Explanation:

We have to dimension

(a) Force

We know that force F = ma

Dimension of mass is M, and dimension of a is [tex]LT^{-2}[/tex]

So dimension of force [tex]MLT^{-2}[/tex]

(B) Energy

We know that energy = force × displacement

In previous part we can see dimension of force is   [tex]MLT^{-2}[/tex]

and dimension of displacement = L

So dimension of energy [tex]ML^2T^{-2}[/tex]

(c) Power

We know that power [tex]p=\frac{energy}{time }[/tex]

In previous part we can see dimension of energy is [tex]ML^2T^{-2}[/tex]

And dimension of time is T

So  dimension of power = [tex]ML^2T^{-3}[/tex]

Answer:

[tex]\Delta T = 28.57°C[/tex]

Explanation:

given data:

temperature at outlet of turbine = 260°C

Flow rate = 2 kg/min = 0.033 kg/s

output power  = 1 kW

specific heat =  1.05 kJ/kg.K

We know that power generated by turbine is equal to change in enthalpy

[tex]W = h_2 - h_1[/tex]

   [tex]= mCp(\Delta T)[/tex]

[tex]1*10^3 = 0.0333*1.05*10^3 * \Delta T[/tex]

[tex]\Delta T = \frac{10^3}{0.033*1.05*10^3}[/tex]

[tex]\Delta T = 28.57°C[/tex]

Answer:

138.9 °C

Explanation:

The datum of quality is saying to us that liquid water is in equilibrium with steam. Saturated water table gives information about this liquid-vapour equilibrium. In figure attached, it can be seen that at 350 kPa of pressure (or 3.5 bar) equilibrium temperature is 138.9 °C

Answer:

The flow is turbulent at both the parts of the tube.

Explanation:

Given:  

Water is flowing in circular tube.  

Inlet diameter is [tex]d_{1}= 2[/tex]m.  

Outlet diameter is [tex]d_{2}= 3[/tex]m.  

Inlet velocity is [tex]V_{1}= 3[/tex] m/s.  

Kinematic viscosity is [tex]\nu =1.24\times  10^{-6}[/tex] m²/s.  

Concept:

Apply continuity equation to find the velocity at outlet.  

Apply Reynolds number equation for flow condition.  

Step1  

Apply continuity equation for outlet velocity as follows:  

[tex]A_{1}V_{1}=A_{2}V_{2}[/tex]  

[tex]\frac{\pi}{4}d^{2}_{1}V_{1}=\frac{\pi}{4}d^{2}_{2}V_{2}[/tex]  

Substitute the values in the above equation as follows:  

[tex]\frac{\pi}{4}2^{2}\times 3=\frac{\pi}{4}3^{2}V_{2}[/tex]  

[tex]2^{2}\times 3=3^{2}V_{2}[/tex]  

[tex]V_{2}=\frac{4}{3}[/tex] m/s.  

Step2

Apply Reynolds number formula for the flow condition at inlet as follows:  

[tex]Re=\frac{v_{1}d_{1}}{\nu }[/tex]  

[tex]Re=\frac{2\times 3}{1.24\times  10^{-6}}[/tex]  

Re=4838709.677  

Apply Reynolds number formula for the flow condition at outlet as follows:  

[tex]Re=\frac{v_{2}d_{2}}{\nu }[/tex]  

[tex]Re=\frac{\frac{4}{3}\times 3}{1.24\times 10^{-6}}[/tex]  

Re=3225806.452  

Thus, the Reynolds number is greater than 2000. Hence the flow is turbulent.  

Hence, the flow is turbulent at both the parts of the tube.  

Answer:

The modulus of toughness is greater for ductile material.

Explanation:

Modulus of toughness is defined as the amount of strain energy that a material stores per unit volume. It is equal to the area under stress-strain curve of the material up to the point of fracture.

As we know that  the area under the stress-strain curve of a ductile material is much more than the area under the stress strain curve of a brittle material as brittle materials fail at a lesser load hence we conclude that the modulus of toughness is greater for ductile material than a brittle material.

In our experience we can also relate that for same volume a substance such as glass sheet (brittle) fails at lower load as compared to a sheet of steel (ductile) with identical dimensions.

Answer:

X1= 41%

heat transfer = -3450.676 KJ

Explanation:

To get the properties for pure substance in a system we need to know at least to properties. These are usually pressure and temperature because they’re easy to measure. In this case we know the initial pressure (20 bar) which is not enough to get all the properties, but they ask to determine quality, this a property that just have meaning in the two-phase region (equilibrium) so with this information we can get the temperature of the system and all its properties.

There is another property that we can calculate from the data. This is the specific volume. This is defined as [tex]\frac{volume}{mass}[/tex]. We know the mass (12 Kg) and we can assume the volume is the volume of the tank  (0.5 [tex]m^{3}[/tex]) because they say that the tank was filled.  

With this we get a specific volume of  

Specific volume = [tex]\frac{0,5 m^{3}}{ 12 kg}= 0.04166667 \frac{m^{3}}{Kg}[/tex]

From the thermodynamic tables we can get the data for the saturated region with a pressure of 20 bar.

Temperature of saturation = 212.385 °C  

Specific volume for the saturated steam (vg) = 0.0995805 [tex]\frac{m^{3}}/{Kg}[\tex]

Specific volume for the saturated liquid (vf)= 0.00117675 [tex]\frac{m^{3}}/{Kg}[\tex]

The specific volume that we calculate before 0.04166667 m^3/Kg is between 0.00117675 m^3/Kg and 0.0995805 m^3/Kg so we can be sure that we are in two-phase region (equilibrium).

The quality (X) is defined as the percentage in mass of saturated steam in a mix (Two-phase region)

The relation between specific volume and quality is  

[tex]v = (1-x)*v_{f} + x*v_{g}[\tex]  

where  

v in the specific volume in the condition (0.04166667 m^3/Kg)  

vf = Specific volume for the saturated liquid (0.00117675 m^3/Kg)

vg = Specific volume for the saturated steam (0.0995805 m^3/Kg)

x = quality

clearing the equation we get:

[tex]X = \frac{(v-v_{f})}{(v_{g}-v_{f})}[/tex] 

[tex]X =\frac{(0.04166667- 0.00117675)}{ (0.0995805 – 0.00117675)} = 0.411[/tex]

The quality is 41%

To calculate the heat transfer we use the next equation.  

Q = m * Cp * delta T  

Where  

Q = heat transfer (Joules, J)

m= mass of the substance (g)

Cp = specific heat (J/g*K) from tables  

Delta T = change in temperature in K for this equation.  

The mass of the substance is 12 kg or 12000 g for this equation  

Cp from tables is 4,1813 J/g*K. You can find this value for water in different states. Here we are using the value for liquid water.  

For delta T, we know the initial temperature 212.385 °C.

We also know that the system was cooled. Since we don’t have more information, we can assume that the system was cooled until a condition where all the steam condensates so now we have a saturated liquid. Since we know the pressure (4 bar), we can get the temperature of saturation for this condition from the thermodynamics tables. This is 143.613 °C, so this is the final temperature for the system.  

T(K) = T°C +273  

T1(K) = 212.385 + 273.15 = 485.535 K

T2 (K) = 143.613 +273.15= 416.763 K

Delta T (K) = (T2-T1) =416.763 K - 485.535 K = -68.772 K

Now we can calculate Q

Q = 12000g * 4,1813 J/g*K* (-68.772 K) = -3450676.36 J or -3450.676 KJ

Is negative because the heat is transfer from the water to the surroundings