Answer:
Explanation:
We are given:
m = 1.06Kg
T = 22kj
Therefore we need to find coefficient performance or the cycle
= 5
For the amount of heat absorbed:
= 5 × 22 = 110KJ
For the amount of heat rejected:
= 110 + 22 = 132KJ
[tex[ q_H = \frac{Q_L}{m} [/tex];
= 
= 124.5KJ
Using refrigerant table at hfg = 124.5KJ/Kg we have 69.5°c
Convert 69.5°c to K we have 342.5K
To find the minimum temperature:
;
= 285.4K
Convert to °C we have 12.4°C
From the refrigerant R -134a table at 
= 12.4°c we have 442KPa
Answer:
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Explanation:
Answer: 24 pA
Explanation:
As pure silicon is a semiconductor, the resistivity value is strongly dependent of temperature, as the main responsible for conductivity, the number of charge carriers (both electrons and holes) does.
Based on these considerations, we found that at room temperature, pure silicon resistivity can be approximated as 2.1. 10⁵ Ω cm.
The resistance R of a given resistor, is expressed by the following formula:
R = ρ L / A
Replacing by the values for resistivity, L and A, we have
R = 2.1. 10⁵ Ω cm. (10⁴ μm/cm). 50 μm/ 0.5 μm2
R = 2.1. 10¹¹ Ω
Assuming that we can apply Ohm´s Law, the current that would pass through this resistor for an applied voltage of 5 V, is as follows:
I = V/R = 5 V / 2.1.10¹¹ Ω = 2.38. 10⁻¹¹ A= 24 pA
Answer:
c
Explanation:
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Answer:
1700kJ/h.K
944.4kJ/h.R
944.4kJ/h.°F
Explanation:
Conversions for different temperature units are below:
1K = 1°C + 273K
1R = T(K) * 1.8
= (1°C + 273) * 1.8
1°F = (1°C * 1.8) + 32
Q/delta T = 1700kJ/h.°C
T (K) = 1700kJ/h.°C
= 1700kJ/K
T (R) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8R
= 944.4kJ/h.R
T (°F) = 1700kJ/h.°C
= 1700kJ/h.°C * 1°C/1.8°F
= 944.4kJ/h.°F
Note that arithmetic operations like subtraction and addition of values do not change or affect the value of a change in temperature (delta T) hence, the arithmetic operations are not reflected in the conversion. Illustration: 5°C - 3°C
= 2°C
(273+5) - (273+3)
= 2 K
