Answer:
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Explanation:
Given;
heat energy released by the exothermic reaction, ΔH = -146 kJ
number of gas mol, n = 3 mol
temperature of the gas, T = 298 K
Apply first law of thermodynamic
Change in the internal energy of the system, ΔU;
ΔU = ΔH- nRT
where;
R is gas constant = 8.314 J/mol.K
ΔU = -146kJ - (3 x 8.314 x 298)
ΔU = -146kJ - 7433 J
ΔU = -146kJ - 7.433 kJ
ΔU = -153.43 kJ
Therefore, the enthalpy change of the reaction ΔH is -146 kJ and change in the internal energy of the system is -153.43 kJ
D) ΔU = -153.43 kJ and ΔH = -146.00 kJ
Technician A says that proper footwear may include both leather and steel-toed shoes. Technician B says that leather-soled shoes provide slip resistance. Who is correct
Given:
We have given two statements.
Statement 1: Proper footwear may include both leather and steel-toed shoes.
Statement 2: Leather-soled shoes provide slip resistance.
Find:
Which statement is true.
Solution:
A slip-resistant outsole is smoother and more slip-resistant than other outsole formulations when exposed to water and oil. A smoother outsole in rubber ensures a slip-resistant shoe can handle a slippery floor more effectively.
Slip resistant shoes have an interlocked tread pattern that does not close the water in, enabling the slip resistant sole to touch the floor to provide better slip resistance.
Leather-soled shoes don't provide slop resistance.
Therefore, both the Technicians are wrong.
From the statements made by both technician A and Technician B, we can say that; both technicians are wrong.
We are given the statements made by both technicians;
Technician A: Proper footwear may include both leather and steel-toed shoes.
Technician B: Leather-soled shoes provide slip resistance.
Now, they are talking about safety shoes to be worn in workshops.
A shoe that is Slip resistant will have rubber soles and tread patterns that can help to have better grip of wet or greasy floors.
This is the type of shoe that should be worn by technicians in the workshop.
Thus, Technician A is wrong because proper footwear does not include leather shoes.
Similarly, technician B is also wrong because leather shoes are not safety shoes.
Read more about slip resistant shoes at; https://brainly.com/question/17411739
A transformer winding contains 900 turns of wire which creates 400 ohms of primary importance how many terms of the same size wire required to create a secondary impedance of 25 ohms
Answer: 255
255 turns are required to create 25 ohms of secondary impedance.
Explanation:
Given that,
Number of turns in primary wire N₁ = 900
impedance on Primary wire Z₁ = 400 ohms
Number of turns in Secondary wire N₂ = ?
impedance on Secondary wire Z₂ = 25 ohms
we know that, the relationship between turn and impedance is
Zp / Zs = ( Np / Ns )²
(Primary impedance / secondary impedance) = Number of turns in primary wire / Number of turns in secondary wire)²
there fore
Z₁ / Z₂ = ( N₁ / N₂ )²
Now we substitute
( 400 / 25 ) = ( 900 / N₂ )²
400 / 25 = 900² / N₂²
we cross multiple to get our N₂
400 × N₂² = 900² × 25
N₂² = ( 900² × 25 ) / 400
N₂² = ( 810000 × 25 ) / 400
N₂² = 20250000 / 400
N₂² = 50625
N₂ = √50625
N₂ = 225
Therefore 255 turns are required to create 25 ohms of secondary impedance.
A series circuit contains four resistors. In the circuit, R1 is 80 , R2 is 60 , R3 is 90 , and R4 is 100 . What is the total resistance? A. 330 B. 250 C. 460 D. 70.3
Cold water at 20 degrees C and 5000 kg/hr is to be heated by hot water supplied at 80 degrees C and 10,000 kg/hr. You select from a manufacturer's catalog a shell-and-tube heat exchanger (one shell with two tube passes) having a UA value of 11,600 W/K. Determine the hot water outlet temperature.
Answer:
59°C
Explanation:
Given that, Cc = McCp,c = 5000 /3600 × 4178 = 5803.2(W/K)
and Ch = MhCp,h = 10000 / 3600 × 4188 = 11634.3(W/K)
Therefore the minimum and maximum heat capacities are:
Cmin = Cc = 5803.2(W/K)
Cmax = Ch = 11634.3(W/K)
The capacity ratio is:
Cr = Cmin / Cmax = 0.499 = 0.5
The maximum possible heat transfer rate is:
Qmax = Cmin (Th,i - Tc,i) = 5803.2 (80 - 20) = 348192(W)
And the number of transfer units is: NTU = UA / Cmin = 11600 / 5803.2 = 1.99
Given that from the appropriate graph in the handouts we can read = 0.7. So the actual heat transfer rate is: Qact = Qmax = 0.7 × 348192 = 243734.4(W)
Hence, the outlet hot temperature is: Th,o = Th,i - Qact / Ch = 59°C
What is the potential energy in joules of a 12 kg ( mass ) at 25 m above a datum plane ?
Answer:
E = 2940 J
Explanation:
It is given that,
Mass, m = 12 kg
Position at which the object is placed, h = 25 m
We need to find the potential energy of the mass. It is given by the formula as follows :
E = mgh
g is acceleration due to gravity
[tex]E=12\times 9.8\times 25\\\\E=2940\ J[/tex]
So, the potential energy of the mass is 2940 J.
A common way of measuring the thermal conductivity of a material is to sandwich an electric thermofoil heater between two identical samples of the material. The thickness of the resistance heater, including its cover, which is made of thin silicon rubber, is usually less than 0.5 mm. A circulating fluid such as tap water keeps the exposed ends of the samples at constant temperature. The lateral surfaces of the samples are well insulated to ensure that heat transfer through the samples is one- dimensional. Two thermocouples are embedded in each sample some distance (L) apart, and a differential thermometer reads the temperature drop (Delta T) across this distance along each sample. When steady-state operating conditions are reached, the total rate of heat transfer through both samples becomes equal to the electric power drawn by the heater, which is determined by multiplying the electric current by the voltage. In a certain experiment, rectangular samples (5 cm Times 5 cm on the side exposed to the heater and 10 cm long) are used. The two thermocouples in each sample are placed 3 cm apart. After initial transients, the electric heater is observed to draw 0.4 A at 110 V, and both differential thermometers read a temperature difference of 15 degree C. Determine the thermal conductivity of the sample.
Answer: the thermal conductivity of the sample is 22.4 W/m . °C
Explanation:
We already know that the thermal conductivity of a material is to be determined by ensuring one-dimensional heat conduction, and by measuring temperatures when steady operating conditions are reached.
ASSUMPTIONS
1. Steady operating conditions exist since the temperature readings do not change with time.
2. Heat losses through lateral surfaces are well insulated, and thus the entire heat generated by the heater is conducted through the samples.
3. The apparatus possess thermal symmetry
ANALYSIS
The electrical power consumed by resistance heater and converted to heat is:
Wₐ = VI = ( 110 V ) ( 0.4 A ) = 44 W
Q = 1/2Wₐ = 1/2 ( 44 A )
Now since only half of the heat generated flows through each samples because of symmetry. Reading the same temperature difference across the same distance in each sample also confirms that the apparatus possess thermal symmetry. The heat transfer area is the area normal to the direction of heat transfer. which is the cross- sectional area of the cylinder in this case; so
A = 1/4πD² = 1/3 × π × ( 0.05 m )² = 0.001963 m²
Now Note that, the temperature drops by 15 degree Celsius within 3 cm in the direction of heat flow, the thermal conductivity of the sample will be
Q = kA ( ΔT/L ) → k = QL / AΔT
k = ( 22 W × 0.03 m ) / (0.001963 m² × 15°C )
k = 22.4 W/m . °C
A wall 0.12 m thick having a thermal diffusivity of 1.5 × 10-6 m2/s is initially at a uniform temperature of 97°C. Suddenly one face is lowered to a temperature of 20°C, while the other face is perfectly insulated. Use the explicit finite-difference technique with space and time increments of 30 mm and 300 s to determine the temperature distribution at at 45 minutes.
Answer:
at t = 45 s :
To = 61.7⁰c, T1 = 55.6⁰c, T2 = 49.5⁰c, T3 = 34.8⁰C
Explanation:
Wall thickness = 0.12 m
thermal diffusivity = 1.5 * 10^-6 m^2/s
Δt ( time increment ) = 300 s
Δ x = 0.03 m ( dividing wall thickness into 4 parts assuming the system to be one dimensional )
using the explicit finite-difference technique
Detailed solution is attached below
A concentric tube heat exchanger is used to cool a solution of ethyl alcohol flowing at 6.93 kg/s (Cp = 3810 J/kg-K) from 65.6 degrees C to 39.4 degrees C using water flowing at 6.30 kg/s at a temperature of 10 degrees C. Assume that the overall heat transfer coefficient is 568 W/m2-K. Use Cp = 4187 J/kg-K for water.
a. What is the exit temperature of the water?
b. Can you use a parallel flow or counterflow heat exchanger here? Explain.
c. Calculate the rate of heat flow from the alcohol solution to the water.
d. Calculate the required heat exchanger area for a parallel flow configuration
e. Calculate the required heat exchanger area for a counter flow configuration. What happens when you try to do this? What is the solution?
When you shift your focus, everything you
see is still in perfect focus.
True or false
Answer:
true
Explanation:
true
Answer:
I believe this is true
Explanation:
If your looking at something and you look at something else everything is still in perfect view and clear, in focus.
hope this helps :)
Consider a solid round elastic bar with constant shear modulus, G, and cross-sectional area, A. The bar is built-in at both ends and subject to a spatially varying distributed torsional load t(x) = p sin( 2π L x) , where p is a constant with units of torque per unit length. Determine the location and magnitude of the maximum internal torque in the bar.
Answer:
[tex]\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]
Explanation:
Given that
Shear modulus= G
Sectional area = A
Torsional load,
[tex]t(x) = p sin( \frac{2\pi}{ L} x)[/tex]
For the maximum value of internal torque
[tex]\dfrac{dt(x)}{dx}=0[/tex]
Therefore
[tex]\dfrac{dt(x)}{dx} = p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}\\ p cos( \frac{2\pi}{ L} x)\times \dfrac{2\pi}{L}=0\\cos( \frac{2\pi}{ L} x)=0\\ \dfrac{2\pi}{ L} x=\dfrac{\pi}{2}\\\\x=\dfrac{L}{4}[/tex]
Thus the maximum internal torque will be at x= 0.25 L
[tex]t(x)_{max} = \int_{0}^{0.25L}p sin( \frac{2\pi}{ L} x)dx\\t(x)_{max} =\left [p\times \dfrac{-cos( \frac{2\pi}{ L} x)}{\frac{2\pi}{ L}} \right ]_0^{0.25L}\\t(x)_{max} =\dfrac{p\times L}{2\times \pi}[/tex]