Question

A 100 w electric bulb was switched on in a 2.5 m 3m 3 m size thermally insulated room having temperature of 20c. The room temperature at the end of 24 hrs. Will be (a) 100c (b) 470c (c) 370c (d) 600c

Answer 1

B is the answer

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Answer 2

Answer: Option B

Explanation: We know the basic formula as

Q=m Cv Dt

First calculate

Q = 100 watt = 100 J/s

For 24 hrs

Q= 100×24×3600 J = 8640KJ

Put value of Q , mass= volume × density , Cv= 0.718 KJ/Kg K and density= 1.2 Kg/cubic.m

8640=1.2×(2.5×3×3)×0.718×(T-20℃)

[ T=445.702℃ ~ 470℃ ]