The milk flows through the heat exchanger at a rate of 1800 kg / hr. The heat exchanger is capable of supplying 111,600 kJ / hr. The final temperature of the product is 80 ° C. Determine the initial temperature of the milk. The specific heat of milk is 3.9 kJ / (kg ° C).
1. The milk flows through the heat exchanger at a rate of 1800 kg / hr. The heat exchanger is capable of supplying 111,600 kJ / hr. The final temperature of the product is 80 ° C. Determine the initial temperature of the milk. The specific heat of milk is 3.9 kJ / (kg ° C).
Jawaban:
D debit susu = 1800 kg/jam
asumsi pada 1 jam, massa susu = 1800 kg.
Q panas yang dibutuhkan 111600 kJ dalam 1 jam
c susu = 3.9 kJ/kg°C
T akhir = 80°C
T awal = ?
Q = m.c.∆T
111600 = 1800 × 3.9 × (80- To)
To = 64.11° C
2. To control inflation rate, the government conducts a fiscal policy which is ... (A) contractionary through increasing tax rate. (B) contractionary through increasing subsidy. (C) contractionary through increasing interest rate. (D) expansionary through increasing subsidy. (E) expansionary through increasing interest rate.
Jawaban:
A
Penjelasan:
pajak naik, uang dari masyarakat berkurang. uang beredar makin dikit. maka nilai uang jadi menguat. nilai uang menguat, maka inflasi teratasi
3. a long aluminum wire of diameter 3 mm is extruded at a temperature of 370°c. the wire is subjected to cross air flow at 30°c at a velocity of 6 m/s. determine the rate of heat transfer from the wire to the air per meter length when it is first exposed to the air.
answer:
Explanation:
Given:
Diameter of aluminum wire, D = 3mm
Temperature of aluminum wire, T_{s}=280^{o}CT
s
=280
o
C
Temperature of air, T_{\infinity}=20^{o}CT
\infinity
=20
o
C
Velocity of air flow V=5.5m/sV=5.5m/s
The film temperature is determined as:
\begin{gathered}T_{f}=\frac{T_{s}-T_{\infinity}}{2}\\\\=\frac{280-20}{2}\\\\=150^{o}C\end{gathered}
T
f
=
2
T
s
−T
\infinity
=
2
280−20
=150
o
C
from the table, properties of air at 1 atm pressure
At T_{f}=150^{o}CT
f
=150
o
C
Thermal conductivity, K = 0.03443 W/m^oCK=0.03443W/m
o
C ; kinematic viscosity v=2.860 \times 10^{-5} m^2/sv=2.860×10
−5
m
2
/s ; Prandtl number Pr=0.70275Pr=0.70275
The reynolds number for the flow is determined as:
\begin{gathered}Re=\frac{VD}{v}\\\\=\frac{5.5 \times(3\times10^{-3})}{2.86\times10^{-5}}\\\\=576.92\end{gathered}
Re=
v
VD
=
2.86×10
−5
5.5×(3×10
−3
)
=576.92
sice the obtained reynolds number is less than 2\times10^52×10
5
, the flow is said to be laminar.
The nusselt number is determined from the relation given by:
Nu_{cyl}= 0.3 + \frac{0.62Re^{0.5}Pr^{\frac{1}{3}}}{[1+(\frac{0.4}{Pr})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{Re}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}Nu
cyl
=0.3+
[1+(
Pr
0.4
)
3
2
]
4
1
0.62Re
0.5
Pr
3
1
[1+(
282000
Re
)
8
5
]
5
4
\begin{gathered}Nu_{cyl}= 0.3 + \frac{0.62(576.92)^{0.5}(0.70275)^{\frac{1}{3}}}{[1+(\frac{0.4}{(0.70275)})^{\frac{2}{3}}]^{\frac{1}{4}}}[1+(\frac{576.92}{282000})^{\frac{5}{8}}]^{\frac{4}{5}}\\\\=12.11\end{gathered}
Nu
cyl
=0.3+
[1+(
(0.70275)
0.4
)
3
2
]
4
1
0.62(576.92)
0.5
(0.70275)
3
1
[1+(
282000
576.92
)
8
5
]
5
4
=12.11
The covective heat transfer coefficient is given by:
Nu_{cyl}=\frac{hD}{k}Nu
cyl
=
k
hD
Rewrite and solve for hh
\begin{gathered}h=\frac{Nu_{cyl}\timesk}{D}\\\\=\frac{12.11\times0.03443}{3\times10^{-3}}\\\\=138.98 W/m^{2}.K\end{gathered}
h=
D
Nu
cyl
\timesk
=
3×10
−3
12.11×0.03443
=138.98W/m
2
.K
The rate of heat transfer from the wire to the air per meter length is determined from the equation is given by:
\begin{gathered}Q=hA_{s}(T_{s}-T{\infin})\\\\=h\times(\pi\timesDL)\times(T_{s}-T{\infinity})\\\\=138.92\times(\pi\times3\times10^{-3}\times1)\times(280-20)\\\\=340.42W/m\end{gathered}
Q=hA
s
(T
s
−T∞)
=h×(π\timesDL)×(T
s
−T\infinity)
=138.92×(π×3×10
−3
×1)×(280−20)
=340.42W/m
The rate of heat transfer from the wire to the air per meter length is Q=340.42W/mQ=340.42W/m
Penjelasan:
.
4. 1. What is Expansion of Heat??
Jawaban:
expansion of shape due to heat (pemuaian)
Penjelasan:
maaf kalo salah
5. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is.
An object with a mass of 200 g has a heat capacity of 104 J/°C. The specific heat of the substance is 20.8 J/kg°C
Solution:
Known:
m = 200 gram = 0,2 kg
C = 104 J/°C
Asked:
c = ...?
Answer:
c = m × c
= 0,2 × 104
= 20.8 J/kg°C
So, the specific heat of the substance is 20.8 J/kg°C.
6. A heated rod with a uniform heat source can be modelled by the Poisson equation d 2θ dx2 = −g(x) where θ is a temperature distribution in the direction of heat flow, x denotes the local position with respect to x–coordinate and g(x) is a heat source. Given g(x) = 28 and the boundary conditions θ(0) = 50 and θ(20) = 400, approximate the temperature distribution using the Finite Difference method. Use a step size, ∆x = 5.
Jawaban:
cari sendiri di buku pasti ada kalu rajin
7. A fish pond of 7.200 l is filled with water at a volumetric flow rate of 30 l/minutes.How long does it take to fill the pond completely?
Jawab:
240min=4 hours
Penjelasan dengan langkah-langkah:
7200Liter:30Liter=240Min+ 4hour
8. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is
Jawaban:
520 J/Kg C
Penjelasan:
diketahui :
m = 200 = 0,2
C = 104 j/kg C
ditanya :
c = ..... ?
dijawab :
c = C/m
= 104/0,2
= 520 j/kg C
9. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is
Jawaban:
520 J/Kg C
diketahui :
m = 200 = 0,2
C = 104 J/C
ditanya :
c = ?
dijawab :
c = C/m
c = 104/0,2
c = 520 J/Kg C
10. an example of conductor of heat is.........
an example of conductor of heat is iron
An example of conductor of heat is copper11. what is the height of a cylindrical tin with radius 14cm and capacity 4804,8 cm³?
[tex]v = \pi \times r \times r \times h \\ 4804,8 = \frac{22}{7} \times 14 \times 14 \times h \\ 4804,8 =616\times h\\ \frac{4804,8}{616} = h \\ h = 7,8 \: cm[/tex]
12. At 0 ° C the sublimation heat of the ice is 675.7 cal / g, while the heat of vaporization of the water is 595.9 cal / g. Calculating the rate of change of pressure steam temperature and ice water at 0 ° C. At this temperature the pressure of water vapor is 4.58 mm Hg.
Jawaban:
The rate of change of pressure with temperature for a substance in a given phase (solid, liquid, or gas) is determined by its vapor pressure curve. The vapor pressure curve shows the pressure of a substance's vapor in equilibrium with its condensed phase (solid or liquid) at a given temperature.
For water, the vapor pressure curve shows that the vapor pressure of water increases as the temperature increases. At 0 °C, the vapor pressure of water is 4.58 mm Hg. As the temperature increases, the vapor pressure of water will also increase, and the rate of change of pressure with temperature will be positive.
The sublimation heat of ice (675.7 cal/g) and the heat of vaporization of water (595.9 cal/g) are both related to the energy required to change the phase of water (from solid to vapor or liquid to vapor, respectively). These values are not directly related to the vapor pressure curve and do not affect the rate of change of pressure with temperature at a given temperature.
Jawaban:
= -37.5 atm/s
Penjelasan:
To calculate the rate of change of pressure for steam and ice water at 0°C, we can use the rate of change of pressure equation, which is defined as the change in pressure per unit time. This equation can be written as follows:
ΔP/Δt = hL(Pv - P)
where:
ΔP is the change in pressure
Δt is the change in time
hL is the rate of change of pressure, which is a constant that depends on the temperature and density of the fluid
Pv is the vapor pressure of the fluid
P is the current pressure of the fluid
To calculate the rate of change of pressure for steam at 0°C, we can use the values given in the question, which are:
hL(steam) = (595.9 cal/g) / (18 g/mol) = 33.1 cal/mol.atm
Pv(steam) = 4.58 mm Hg = 0.006 atm
P(steam) = 1 atm (standard atmospheric pressure at 0°C)
Using the above equation, we can calculate the rate of change of pressure for steam at 0°C as follows:
ΔP/Δt = hL(Pv - P)
= (33.1 cal/mol.atm) * (0.006 atm - 1 atm)
= -0.198 atm/s
On the other hand, to calculate the rate of change of pressure for ice water at 0°C, we can use the values given in the question, which are:
hL(ice water) = (675.7 cal/g) / (18 g/mol) = 37.5 cal/mol.atm
Pv(ice water) = 0 atm (the vapor pressure of ice is 0 at 0°C)
P(ice water) = 1 atm (standard atmospheric pressure at 0°C)
Using the above equation, we can calculate the rate of change of pressure for ice water at 0°C as follows:
ΔP/Δt = hL(Pv - P)
= (37.5 cal/mol.atm) * (0 atm - 1 atm)
= -37.5 atm/s
13. A certain insulation has a thermal conduktivity of 10 W/m. drjtC. what thicknrss is necessary to offect a temperature drop of 5 drajat C for a heat flow of 400 W/m².
Di atas adalah rumus yang akan kita gunakan dengan H adalah kalor yang merambat tiap satuan waktu.
Diketahui :
k = 10 W/mC
ΔT = -5°C (turun 5°)
h = 400 W/m^2
Ditanya: Ketebalan (l)
Jawab :
h = -k. A. ΔT/l
400 = (-10) (-5)/l
l = 50/400
l = 1/8 = 1,25 m
14. heat can be transfered through a vacuum by
... radiation, just like how the heat got transfered from the sun through the earth
15. An object with a mass of 200 g has a heat capacity of 104 J/oC. The specific heat of the substance is
An object with a mass of 200 g has a heat capacity of 104 J/°C. The specific heat of the substance is 20.8J/kg°C
Solution:Known:
m = 200 gram = 0,2 kg
C = 104 J/°C
Asked:
c = ...?
Answer:
c = m × c
= 0,2 × 104
= 20.8J/kg°C
So, the specific heat of the substance is 20.8 J/kg°C.
16. what is the comparative of "heat"?
Jawaban:
apa perbandingan "panas"?
jawaban=
Penjelasan:
Panas, bahang, atau kalor adalah energi yang berpindah akibat perbedaan suhu. Satuan SI untuk panas adalah joule.
Panas bergerak dari daerah bersuhu tinggi ke daerah bersuhu rendah.[1][2] Setiap benda memiliki energi dalam yang berhubungan dengan gerak acak dari atom-atom atau molekul penyusunnya.
Energi dalam ini berbanding lurus terhadap suhu benda. Ketika dua benda dengan suhu berbeda bergandengan, mereka akan bertukar energi internal sampai suhu kedua benda tersebut seimbang. Jumlah energi yang disalurkan adalah jumlah energi yang tertukar. Kesalahan umum untuk menyamakan panas dan energi internal. Perbedaannya adalah panas dihubungkan dengan pertukaran energi internal dan kerja yang dilakukan oleh sistem. Mengerti perbedaan ini dibutuhkan untuk mengerti hukum pertama termodinamika.
Radiasi inframerah sering dihubungkan dengan panas, karena objek dalam suhu ruangan atau di atasnya akan memancarkan radiasi kebanyakan terpusat pada rentang inframerah-tengah (lihat badan hitam).
Jawaban:
cold
jadikanjawaban paling terbaik
17. The water flow rate is 80 liters/minute. If the water flows for 1 hour, then the volume of the water is . . . . liter.
Jawaban:
4800 liter bcs 80 liter x 60 minute
Jawaban:
4.800 liters
Penjelasan dengan langkah-langkah:
80 liters/minute = 80 liters ÷ 1/60 hour = 4.800 liters/hour
Volume = water discharge × time
= 4.800 liters/hour × 1 hour
= 4.800 liters
18. 5. The volume of water tub in bathroom is 500 liters. The flow rate of the water tap is 25cm³/minute. How long that the water tub will be full of water?
Materi : Bangun Ruang dan Volume
Volume = 500 L = 500 dm³
Debit = 25 cm³/menit
\_____________________/
Waktu
= 500.000 cm³/ 25 cm³ × 1 menit
= 20.000 × 1 menit
= 20.000 menit
= 333 jam 20 menit
= 13 hari 21 jam 20 menit
Semoga bisa membantu
[tex] \boxed{ \colorbox{navy}{ \sf{ \color{lightblue}{ Answer\:by\: BLUEBRAXGEOMETRY}}}} [/tex]
19. 5. Electric current cannot only flow through an _______ circuit.6. Electric current can only flow through a ________ circuit
Jawaban:
5.closed
6.If the electric circuit is in a closed state
Penjelasan:
KASIH ROLE JAWABAN TERCERDAS YA
20. T he forming section of a plastics plant puts out a continuous sheet of plastic that is 1.2 m wide and 2 mm thick at a rate of 15 m/min. The temperature of the plastic sheet is 90°C when it is exposed to the surrounding air, and the sheet is subjected to air flow at 30°C at a velocity of 3 m/s on both sides along its surfaces normal to the direction of motion of the sheet. The width of the air cooling section is such that a fixed point on the plastic sheet passes through that section in 2 s. Determine the rate of heat transfer from the plastic sheet to the air.
Jawaban:
m/s on both sides along its surfaces normal to
is exposed
0 Komentar