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JEE Main 2020 · Physics · Calorimetry and Heat Transfer · medium

Steam at 100^circC is passed into a vessel of water equivalent 10 g containing 44 g of ice, both initially at -10^circC, until the contents reach…

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Steam at 100∘C100^\circ\mathrm{C} is passed into a vessel of water equivalent 10 g10\ \mathrm{g} containing 44 g44\ \mathrm{g} of ice, both initially at −10∘C-10^\circ\mathrm{C}, until the contents reach 40∘C40^\circ\mathrm{C}. Taking sice=0.5s_{ice} = 0.5, Lf=80L_f = 80, Lv=540 cal g−1L_v = 540\ \mathrm{cal\,g^{-1}} and swater=1 cal g−1 ∘C−1s_{water} = 1\ \mathrm{cal\,g^{-1}\,{}^\circ C^{-1}}, the mass of steam condensed is
  1. A. 8 g8\ \mathrm{g}
  2. B. 12 g12\ \mathrm{g}
  3. C. 10 g10\ \mathrm{g}
  4. D. 9.6 g9.6\ \mathrm{g}
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Answer: (C) 10 g10\ \mathrm{g}

Heat gained: ice warms to 0∘C0^\circ\mathrm{C}: 44×0.5×10=22044\times0.5\times10 = 220 cal; melts: 44×80=352044\times80 = 3520 cal; the water warms to 40∘C40^\circ\mathrm{C}: 44×40=176044\times40 = 1760 cal; the vessel warms through 50∘C50^\circ\mathrm{C}: 10×50=50010\times50 = 500 cal. Total =6000= 6000 cal. Each gram of steam gives 540+1×(100−40)=600540 + 1\times(100-40) = 600 cal. Hence m=6000/600=10 gm = 6000/600 = 10\ \mathrm{g}. Forgetting the vessel gives 5500/600≈9.2 g5500/600 \approx 9.2\ \mathrm{g}.
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