Q.

Air (sp. Heat at room temperature and 1 atm=0.71 J/g K ; density) =1.22×10-6 kg cm3 present in a room (dimensions 6×5×4m3 ) got heated up when a bulb of 500 watt was switched on for 10 minutes, the increase in room temperature if heat capacity of four walls and roof is 60×103J/K is:

see full answer

Want to Fund your own JEE / NEET / Foundation preparation ??

Take the SCORE scholarship exam from home and compete for scholarships worth ₹1 crore!*
An Intiative by Sri Chaitanya

a

183 K

b

1.83 K

c

213 K

d

715 K

answer is B.

(Unlock A.I Detailed Solution for FREE)

Ready to Test Your Skills?

Check your Performance Today with our Free Mock Test used by Toppers!

Take Free Test

Detailed Solution

1 Watt=1 J/s ; volume of room =6×5×4m3=120 m3=120×106 cm3

Wt. of air in room= volume x density=120×106 cm3×1.22×10-6 kg cm-3

=120×106×1.22×10-6×103g=146400g

Heat given out by bulb=500 watt x (10 x 60) s

                                                =500 J/s x 600s

=3×105J  [1 watt =1J/s]s

 Heat gained by (walls+roof) =60×103 J/K(ΔT)

 Heat gained by air =0.71Jg-K-×146400 g×ΔT

But heat given out by bulb= Heat gained by (walls+roof)+Heat gained by air

3×105J=60×103JK-(ΔT)+0.71×146400JK-×ΔT 3×105J=ΔT60×103+0.71×146400JK-=163944JK-×ΔT ΔT=3×105J163944JK=1.83K

Watch 3-min video & get full concept clarity
AITS_Test_Package
AITS_Test_Package
score_test_img

Get Expert Academic Guidance – Connect with a Counselor Today!

whats app icon