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nadezda [96]
3 years ago
5

What is the lowest Temperature in degrees C?, In degrees K? in degrees F? in degrees R

Engineering
1 answer:
Colt1911 [192]3 years ago
3 0

Answer:

-273.16 °C

-459.677 °F

0 °K

0 °R

Explanation:

The lowest temperature is the absolute zero.

Absolute zero is at 0 degrees Kelvin, or 0 degrees Rankine, because these are absolute scales that have their zero precisely at the absolute zero.

Celsius and Fahrenheit degrees are relative scales, these have their zeroes above the absolute zero.

Celsius scale has the same degree separation as the Kelvin scale, but the zero is separated by 273.16 degrees. Therefore the lowest temperature in the Celsius scale is -273.16 °C.

The Fahrenheit degrees have the same degree separation as the Rankine degrees, and the zero is 459.67 degrees. Therefore the lowest temperature in the Fahrenheit scale is -459.67 °F.

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3 years ago
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The structure of a house is such that it loses heat at a rate of 4500kJ/h per °C difference between the indoors and outdoors. A
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Answer:

15.24°C

Explanation:

The quality of any heat pump pumping heat from cold to hot place is determined by its coefficient of performance (COP) defined as

COP=\frac{Q_{in}}{W}

Where Q_{in} is heat delivered into the hot place, in this case, the house, and W is the work used to pump heat

You can think of this quantity as similar to heat engine's efficiency

In our case, the COP of our heater is

COP_{heater} = \frac{\frac{4500\ kJ}{3600\ s} *(T_{house}-T_{out})}{4\ kW}

Where T_{house} = 24°C and T_{out} is temperature outside

To achieve maximum heating, we will have to use the most efficient heat pump, and, according to the second law of thermodynamics, nothing is more efficient that Carnot Heat Pump

Which has COP of:

COP_{carnot}=\frac{T_{house}}{T_{house}-T_{out}}

So we equate the COP of our heater with COP of Carnot heater

\frac{1.25 *(T_{house}-T_{out})}{4}=\frac{T_{house}}{T_{house}-T_{out}}

Rearrange the equation

\frac{1.25}{4}(24-T_{out})^2-24=0

Solve this simple quadratic equation, and you should get that the lowest outdoor temperature that could still allow heat to be pumped into your house would be

15.24°C

4 0
3 years ago
An excavation is at risk for cave-in and water accumulation because of the excess soil that has accumulated. What type of excava
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Answer:

Among the different types of excavation protection system, as a way of preventing accidents against cave-ins, the sloping involves cutting back the trench wall at an angle inclined away from the excavation. Shoring requires installing aluminum hydraulic or other types of supports to prevent soil movement and cave-ins. Shielding protects workers by using trench boxes or other types of supports to prevent soil cave-ins (OSHA). In addition, the regulations do not allow employees to work on excavations where there is an accumulation of water. If this occurs, water on the site must be constantly removed by suitable equipment preventing water from accumulating. The entry of surface water into the excavations must also be prevented by means of diversion ditches, dam, or other suitable means.  

Explanation:

3 0
3 years ago
1. A 260 ft (79.25 m) length of size 4 AWG uncoated copper wire operating at a tem-
Murljashka [212]

A 260 ft (79.25m) length of size 4 AWG uncoated copper wire operating at a temperature of 75°c has a resistance of 0.0792 ohm.

Explanation:

From the given data the area of size 4 AWG of the code is 21.2 mm², then K is the Resistivity of the material at 75°c is taken as ( 0.0214 ohm mm²/m ).

To find the resistance of 260 ft (79.25 m) of size 4 AWG,

R= K * L/ A

K = 0.0214 ohm mm²/m

L = 79.25 m

A = 21.2 mm²

R = 0.0214 * \frac{79.25}{21.2}

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Thus the resistance of uncoated copper wire is 0.0792 ohm

5 0
3 years ago
1. An air standard cycle is executed within a closed piston-cylinder system and consists of three processes as follows:1-2 = con
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Answer:

Explanation: Here it is: 67 Hope that helps! :)

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