Solution :
Given :
Initial temperature of the refrigerant is :
![$T_i=39.37 ^ \circ C$](https://tex.z-dn.net/?f=%24T_i%3D39.37%20%5E%20%5Ccirc%20C%24)
= ( 39.37 + 273 ) K
= 312.3 K
Room which is maintained at constant temperature is :
![$T_o=22 ^ \circ C$](https://tex.z-dn.net/?f=%24T_o%3D22%20%5E%20%5Ccirc%20C%24)
= (22+273) K
= 295 K
The thermal energy transferred to the room is :
Q = 400 kJ
= ![$400 \times 10^3 \ J$](https://tex.z-dn.net/?f=%24400%20%5Ctimes%2010%5E3%20%5C%20J%24)
Therefore, the total entropy generation during the thermal energy process is :
![$\Delta S =\left[\frac{-Q}{T_i}+ \frac{+Q}{T_i}\right]$](https://tex.z-dn.net/?f=%24%5CDelta%20S%20%3D%5Cleft%5B%5Cfrac%7B-Q%7D%7BT_i%7D%2B%20%5Cfrac%7B%2BQ%7D%7BT_i%7D%5Cright%5D%24)
Here, -Q = heat is leaving the system maintained at a temperature of
K.
+Q = heat is entering the system maintained at a temperature of
K.
Therefore, substituting the values :
![$\Delta S =\left[\frac{-400\times 10^3}{312.3}+ \frac{400\times 10^3}{295}\right]$](https://tex.z-dn.net/?f=%24%5CDelta%20S%20%3D%5Cleft%5B%5Cfrac%7B-400%5Ctimes%2010%5E3%7D%7B312.3%7D%2B%20%5Cfrac%7B400%5Ctimes%2010%5E3%7D%7B295%7D%5Cright%5D%24)
= [-1280.8197 + 1355.9322]
= 75.1125 J/K
= 0.0751125 kJ/K
= 0.075 kJ/K
Sn(s) + 2 HF(aq) ----> SnF2(s) + H2(g)
M HF = 20 g/mol
M SnF2 = 157 g/mol
40 g (HF) -----> 157 g (SnF2)
30 g (HF) -----> x g (SnF2)
x = 30 g × 157 g / 40 g = 117,75g
Answer: 117,75 g
:-) ;-)
Answer:
In simple terms, the endothermic reactions absorb energy from the surrounding that is in the form of heat. On the other hand, an exothermic reaction releases energy into the surrounding of the system.
Odorants such as Citronellal, Linalool, Nerolidol