Assume final temperature =T∘C Heat lass = Heat gain =msΔT ⇒mBsBΔTB=mwswΔTw 0.1×400×(500−T) =0.5×4200×(T−30)+800( T−30) ⇒40(500−T)=(T−30)(2100+800) ⇒20000−40 T=2900 T−30×2900 ⇒20000+30×2900=T(2940) T=30.4∘C TΔT×100=306.4×100=21% so the closest answer is 20%.
A metal ball of mass 0.1 ~kg is heated upto 500° C and dropped into a vessel of heat capacity 800 JK^-1 and containing 0.5 ~kg water. The initial…
JEE Main 2019 — Physics Thermodynamics
A metal ball of mass 0.1 kg is heated upto 500∘C and dropped into a vessel of heat capacity 800JK−1 and containing 0.5 kg water. The initial temperature of water and vessel is 30∘C. What is the approximate percentage increment in the temperature of the water? [Specific Heat Capacities of water and metal are, respectively, 4200Jkg− 1 K−1 and 400Jkg−1 K−1 ]
Held on 11 Jan 2019 · Verified 6 Jul 2026.
Options
- A
15%
- B
30%
- C
25%
- D
20%
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