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JEE Main 201910 JanuaryEvening Shift
Exam
JEE Main
Year
2019
Date
10 January
Shift
Evening Shift
Subject
Physics
Chapter
Kinetic Theory of Gases

JEE Main 2019 · 10 January · Evening Shift · Physics · Kinetic Theory of Gases

An unknown metal of mass 192g heated to a temperature of 100 ° C...

An unknown metal of mass 192g192g heated to a temperature of 100°C100 ° C was immersed into a brass calorimeter of mass 128g128g containing 240g240g of water at a temperature of 8.4oC.{8.4}^{o}C . Calculate the specific heat of the unknown metal if water temperature stabilizes at 21.5oC.{21.5}^{o}C . (( Specific heat of brass is 394Jkg−1K−1)394Jk{g}^{- 1}{K}^{- 1} )

Options

  1. A916Jkg−1K−1916Jk{g}^{- 1}{K}^{- 1}
  2. B458Jkg−1K−1458Jk{g}^{- 1}{K}^{- 1}
  3. C654Jkg−1K−1654Jk{g}^{- 1}{K}^{- 1}
  4. D1232Jkg−1K−11232Jk{g}^{- 1}{K}^{- 1}

Correct answer

A. 916Jkg−1K−1916Jk{g}^{- 1}{K}^{- 1}

Step-by-step solution

Conservation of energy:

ms\DeltaT=(mbsb+mwsw)ΔTwms\DeltaT = \left({m}_{b}{s}_{b} + {m}_{w}{s}_{w}\right)\Delta{T}_{w}

192×10−3s(100−21.5)=(128×394+240×4182)×10−3×(21.5−8.4)192 \times {10}^{- 3}s\left(100 - 21.5\right) = \left(128 \times 394 + 240 \times 4182\right) \times {10}^{- 3} \times \left(21.5 - 8.4\right)

s=(50432+1003680)(13.1)15072=916.19≃916s = \dfrac{\left(50432 + 1003680\right)\left(13.1\right)}{15072} = 916 . 19 ≃ 916 Jkg−1K−1Jk{g}^{- 1}{K}^{- 1}
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