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:
A worn inner CV joint often makes a clunking noise during starts and stops.
Answer:
The feature to configure is Isolation Option which can either be Guest network or Wireless Isolation
Create a guest ssid separate from the Internal Wi-Fi used at home. The guest ssid ensures that the parents have a separate point to the internet from the teacher.
Guset SSID ensure path isolation of guest traffic from her personal data traffic
Explanation:
Answer:
(a) T₂ =747.5 and K= 474.5 °C (b) 330.178 kJ/kg
Explanation:
Solution
T₁ = 35°C = 308
the first step to take is to Use the Table A-17: Ideal gas properties for air:
Now,
At T₁ = 308 K
V₁ = 217.67 + [308-305/310-305] (221.25 -217.67)
So,
V₁ =219.818 kJ/kg
Thus,
Vr₁ = 596 + [308-305/310-305] (572.3 - 596)
= 581.78
so,
Vr₂/Vr₁ = 1/10
Vr₂ =58.178
Applying Table A-17, at Vr₂ = 58.178
Then,
(a) T₂ = 740 + [58.178 - 59.82/57.63 -59.82] (750 -740)
T₂ = 747.5 and K = 474.5 °C
V₂ =544.02 + [58.178 - 59.82/57.63 -59.82] (551.99 - 544.02)
so,
V₂ =549.996 kJ/kg
Hence,
(b) q - w = v₂ -v₁
= 0 -w = 549.996- 219.818
w = 330.178 kJ/kg