Request a call back

Join NOW to get access to exclusive study material for best results

JEE Class main Answered

100 g of water is heated from 30°C to 50°C.Ignoring the slight expansion of the water ,the change in its internal energy is (Take,specific heat of water is 4184 J/Kg/K).     Personally I want to know what is slight expansion and anotherly is the work done here is zero ,if yes how??
Asked by vishakhachandan026 | 02 Apr, 2019, 20:09: PM
answered-by-expert Expert Answer
expansion coefficient of water γ = 2×10-4 / °C
 
hence change in volume = V0 γ ΔT = 100× 2×10-4 × 20 = 0.4 cc  
 
hence change in volume is from 100 cc to 100.4 cc
 
First law of thermodynamics :  ΔQ = ΔU + ΔW  .....................(1)
 
where ΔQ = heat energy given to the system ( here system is 100 g water )
ΔU = change in internal energy,
ΔW = workdone  = pΔV  , where p is pressure and ΔV is change in volume
 
since ΔV≈0, workdone is zero.
 
hence ΔQ = ΔU = m×Cp ×ΔT  = 100×10-3 ×4184×20 = 8368 J
Answered by Thiyagarajan K | 02 Apr, 2019, 20:51: PM
JEE main - Physics
Asked by rambabunaidu4455 | 03 Oct, 2024, 16:03: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by ratchanavalli07 | 17 Sep, 2024, 07:46: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by adithireddy999 | 03 Sep, 2024, 09:35: AM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vaishalinirmal739 | 29 Aug, 2024, 18:07: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT
JEE main - Physics
Asked by vradhysyam | 26 Aug, 2024, 17:17: PM
ANSWERED BY EXPERT ANSWERED BY EXPERT