Thermal radiation is a form of heat transfer because the electromagnetic radiation emitted from the source carries energy away from the source to surrounding (or distant) objects.
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Answer:
COP of heat pump=3.013
COP of cycle=1.124
Explanation
W = Q2 - Q1 ----- equation 1
W = work done
Q2 = final energy
Q1 = initial energy
A) calculate the COP of the heat pump
COP =Q2/W
from equation 1
Q2 = Q1 + W = 15 + 7.45 = 22.45 KW
therefore COP =22.45/7.45 = 3.013
B) COP when cycle is reversed
COP = Q1/W
from equation 1
Q1 + W = Q2 ------ equation 2
Q2 = 15 Btu/s = 15 * 1.055 = 15.825 KW therefore from equation 2
Q1 = 8.375 KW
COP =8.375/7.45 = 1.124
Answer:
The final temperature of water is 381.39 °C.
Explanation:
Given that
Mass of water = 5 kg
Heat transfer at constant pressure Q = 2960 KJ
Initial temperature = 240 °C
We know that heat transfer at constant pressure given as follows

We know that for water

Lets take final temperature of water is T
So


T=381.39 °C
So the final temperature of water is 381.39 °C.
Answer:
a) Please see attached copy below
b) 0.39KJ
c) 20.9‰
Explanation:
The three process of an air-standard cycle are described.
Assumptions
1. The air-standard assumptions are applicable.
2. Kinetic and potential energy negligible.
3. Air in an ideal gas with a constant specific heats.
Properties:
The properties of air are gotten from the steam table.
b) T₁=290K ⇒ u₁=206.91 kj/kg, h₁=290.16 kj/kg.
P₂V₂/T₂=P₁V₁/T₁⇒ T₂=P₂T₁/P₁ = 380/95(290K)= 1160K
T₃=T₂(P₃/P₂)⁽k₋1⁾/k =(1160K)(95/380)⁽⁰°⁴/₁.₄⁾ =780.6K
Qin=m(u₂₋u₁)=mCv(T₂-T₁)
=0.003kg×(0.718kj/kg.k)(1160-290)K= 1.87KJ
Qout=m(h₃₋h₁)=mCp(T₃₋T₁)
=0.003KG×(1.005kj/kg.k(780.6-290)K= 1.48KJ
Wnet, out= Qin-Qout = (1.87-1.48)KJ =0.39KJ
c)ηth= Wnet/W₍in₎ =0.39KJ/1.87KJ = 20.9‰
Answer:
A,C, and D
Explanation:
Potible ladders have to configure with many designs in mind but the most evedent is that they are usally unstable
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