Answer :
Answer:
Q=1005 J
t= 0.67 sec
Explanation:
Lets take condition of room is 1 atm and 25°C.
Heat capacity ,c = 21 J /K.mol
If we assume that air is ideal gas that
P V = n R T
[tex]V= 5.5\times 6.5\times 3\ m^3[/tex]
[tex]V=107.25\ m^3[/tex]
[tex]V=107.25\times 1000 L[/tex]
V= 107250 L
At STP number of moles given as
[tex]n=\dfrac{V}{V_{at\ S.T.P.}}[/tex]
V=22.4 L at S.T.P.
[tex]n=\dfrac{107250}{22.4}\ moles[/tex]
n=4787.94 moles
n= 4.784 Kmoles
So heat required to raise 10°C temperature
Q = n x c x ΔT
Q = 4.78794 x 21 x 10
Q=1004.64 J
Time t
t= Q/P
P= 1.5 KW
t = 1.004.64 /1.5
t= 0.66 sec
Answer:
The energy is 2.201x10⁴kJ and the time is 1.467x10⁴ s
Explanation:
please, the solution is in the attached Word file