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(a) A 1000 watt hour is kept on ina room whose dimensions are 5m xx 5m xx 3m. Howmuch temperature of the room will rise after half an hour? Given the following data : Heat capacityofair at room temperature and 1 atm= 0.71 Jg^(-1)K^(-1) Density ofair = 1.22xx 10^(-3) g mL^(-1) (b) How much temperature of the room willrise if 25% heat is lost by radiation ?

Answer» <html><body><p></p>Solution :(a) 1 Watt `= 1 J s^(-1)` <br/> `:. `Heat evolved by 1000 W heaterin <a href="https://interviewquestions.tuteehub.com/tag/30min-1853364" style="font-weight:bold;" target="_blank" title="Click to know more about 30MIN">30MIN</a>. `= 100 xx 30 xx 60 J = 18 xx 10^(5) J` <br/> Volume of the room ` 5 xx4 xx 3m^(3) = 60 m^(3) = 60xx 10^(6) cm^(3) ` <br/> Density of air`=1.22 xx 10^(-3) g mL^(-1)` <br/> Massof air in the room`= ( 60xx 10^(6)) xx ( 1.22 xx 10^(-3)) g = 73.2 xx 10^(3)g` <br/> Heat evolved by the heater `=` Heat absorbed by <a href="https://interviewquestions.tuteehub.com/tag/walls-1448732" style="font-weight:bold;" target="_blank" title="Click to know more about WALLS">WALLS</a>, roof and air <a href="https://interviewquestions.tuteehub.com/tag/inside-1045864" style="font-weight:bold;" target="_blank" title="Click to know more about INSIDE">INSIDE</a>. <br/> `18 xx10^(5) J =( 5 xx 10^(4) xx Delta T ) +( 73.2 xx 10^(3)) xx 0.71 xx Delta T` <br/> `= ( 50000+ 51972 )Delta T` <br/> `= 101972 Delta T = 1.01 <a href="https://interviewquestions.tuteehub.com/tag/972-342760" style="font-weight:bold;" target="_blank" title="Click to know more about 972">972</a> xx 10^(5) DeltaT ` <br/>or`Delta T = ( 1.8 xx 10^(5))/( 1.01972 xx 10^(5)) = 17.65^(@) ` <br/> (b) If `25%` heat is lost by radiation, net heat available in the room <br/> ` = (75)/( 100) xx 18 xx 10^(5) J = 13.5 xx 10^(5) J` ,<a href="https://interviewquestions.tuteehub.com/tag/brgt-2487365" style="font-weight:bold;" target="_blank" title="Click to know more about BRGT">BRGT</a> `13.5 xx 10^(5) J = ( 5 xx 10^(4) xx Delta T ) + ( 73.2 xx 10^(3)) xx 0.71xx DeltaT` <br/> or`Delta T =( 13.5 xx 10^(5))/( 1.01972xx 10^(5)) = 13.24^(@)`</body></html>


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