Answer:
a)
b)
c) 0 J/K
d)S= 61.53 J/K
Explanation:
Given that
T₁ = 745 K
T₂ = 101 K
Q= 7190 J
a)
The entropy change of reservoir 745 K
Negative sign because heat is leaving.
b)
The entropy change of reservoir 101 K
c)
The entropy change of the rod will be zero.
d)
The entropy change of the system
S= S₁ + S₂
S = 71.18 - 9.65 J/K
S= 61.53 J/K
Answer:
(a) Heat transfer to the environment is: 1 MJ and (b) The efficiency of the engine is: 41.5%
Explanation:
Using the formula that relate heat and work from the thermodynamic theory as:
solving to Q_out we get:
this is the heat out of the cycle or engine, so it will be heat transfer to the environment. The thermal efficiency of a Carnot cycle gives us: 
where T_Low is the lowest cycle temperature and T_High the highest, we need to remember that a Carnot cycle depends only on the absolute temperatures, if you remember the convertion of K=°C+273.15 so T_Low=150+273.15=423.15 K and T_High=450+273.15=723.15K and replacing the values in the equation we get:
I believe flowing water changes the land because eventually frozen water has to melt to flowing water, right?
Answer:
BELOW
Explanation:
Seattle (near water) would have hotter temperatures because it is by a large body of water. Unlike Bismarck which is surrounded by land. Bismarck would get hotter. Seattle would be affected the large body of water which heats more slowly than the continent warming.
