Questions
A 10 W electric heater is used to heat a container filled with 0.5 kg of water. It is found that the temperature of water and container rises by 3K in 15 min. The container is then emptied, dried and filled with 2 kg of oil. The same heater now raises the temperature of container oil system by 2K in 20 min. Assume there is no heat loss in the process and the specific heat of water is 4200 J kg-1K-1 , the specific heat of oil in the same limit is equal to
detailed solution
Correct option is B
As we know, work done by an electric heater i.e. m1c1Δt+m2c2Δt= work done So, m1c1Δt+m2c2Δt=P1t1where, m1=0.5kg, specific heat, c1=4200Jkg−1K−1Δt=Δt1=Δt2=3KP1=10Wt1=15×60=900sc2 = specific heat capacity of container. So, 0.5×4200×(3−0)+m2c2×(3−0)=10×15×60⇒ 2100×3+m2c2×3=9000⇒ m2c2=9000−63003=900Similarly, in the case of oil, m1coΔt+m2c2Δt=P2t2 where, co= specific heat capacity of oil, P2=10W2×c0×2+900×2=10×20×60⇒ 4c0+1800=12000⇒ co=2550=2.55×103Jkg−1K−1Talk to our academic expert!
Similar Questions
Specific heat of a substance can be:
(i) finite
(ii) infinite
(iii) zero
(iv) negative
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