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Question: Answered & Verified by Expert
A calorimeter of negligible heat capacity contains 100 g water at 40 °C. The water cools to 35 °in min. If water is now replaced by a liquid of the same volume as that of water at the same initial temperature it cools to 35 °C in 2 min. Given specific heats of water and liquid are 4200 J kg-1 °C-1 and 2100 J kg-1 °C-1 respectively. Find the density of liquid give the answer in kg m-3. [Assume Newton law of cooling is applicable]
PhysicsThermal Properties of MatterJEE Main
Solution:
2891 Upvotes Verified Answer
The correct answer is: 8
-dTdt=K100×SwT-T0
40 35 dT T T 0 = 0 5 K 100× S w dt
40 35 dT T T 0 = 0 2 dt 100× ρ l s l
5K100Sw=2K100×ρlsl
ρ l = 4 5 g/ cm 3 = 4× 10 3 kg 5× 10 6 m 3 = 4 5 × 10 3
= 40 5 × 10 2 =800kg/ m 3

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