Conservation of energy:
msΔT=(mbsb+mwsw)ΔTw
192×10−3s(100−21.5)=(128×394+240×4182)×10−3×(21.5−8.4)
s=15072(50432+1003680)(13.1)=916.19≃916 Jkg−1K−1
JEE Main 2019 — Physics Thermodynamics
An unknown metal of mass 192g heated to a temperature of 100∘C was immersed into a brass calorimeter of mass 128g containing 240g of water at a temperature of 8.4oC. Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC. ( Specific heat of brass is 394Jkg−1K−1)
Held on 10 Jan 2019 · Verified 6 Jul 2026.
916Jkg−1K−1
458Jkg−1K−1
654Jkg−1K−1
1232Jkg−1K−1
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