Answer:
[tex]Q=9.2x10^5J[/tex]
[tex]t=614s=10.2min[/tex]
Explanation:
Hello,
In this case, we can compute the energy by using the following formula for air:
[tex]Q=nCp\Delta T[/tex]
Whereas the moles of air are computed via the ideal gas equation at room temperature inside the 5.5m x 6.5m x 3.0m-room:
[tex]n=\frac{PV}{RT}\\\\V=5.5m*6.5m*3.0m=107.25m^3*\frac{1000L}{1m^3}=107250L\\ \\n=\frac{1atm*107250L}{0.082\frac{atm*L}{mol*K}*298.15K}\\ \\n=4386.8mol[/tex]
Now, we are able to compute heat, by considering that the temperature raise is given in degree Celsius or Kelvins as well:
[tex]Q=4386.8mol*21\frac{K}{mol*K}*10K \\\\Q=9.2x10^5J[/tex]
Finally, we compute the time required for the heating by considering the heating rate and the required heat, shown below:
[tex]t=\frac{9.2x10^5J}{1.5\frac{kJ}{s}*\frac{1000J}{1kJ} } \\\\t=614s=10.2min[/tex]
Regards.