Question

1.8- An unknown mass of iron at 90C ° is dropped into an insulated tank that contains 8OL of water at 20C °. At the same time, a paddle wheel driven by a 200W motor is activated to stir the water. Thermal equilibrium is established after 25 minutes with a final temperature of 270 ° By neglecting the energy stored in the paddle wheel, determine the mass of the iron.

Answer 1

Let heat carrying capacity of iron block (m × cp) = C

Tfe = 85°C = 358 K

Tw = 20°C = 293 K;

Tf = 24°C = 297 K

Mass of water = 0.1×1000 = 100 kg

W.D by motor in 20 min = 200×20×60 = 240000 J

Heat gain by water = W.D by motor + heat lost by

iron block

Þ 100×4180×(297 – 293) = 240000 + C×(358 – 297)

Þ C = 23475.41 J/K

ΔSwater = 100 × 4.18 × ln(297/293)=5.6679 KJ/K

ΔSFe=23.475xln(297/358)=-4.385 KJ/K

ΔSpadde = 0 because work supplied to a viscous thermally insulated liquid, which is dissipated adiabatically into internal energy increases of liquid, the temperature of which increases from Ti to Tt

\ ΔSuniverse = 5.6679 – 4.3852 + 0

= 1.2827 KJ / K

hope this helps buddy

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