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mr Goodwill [35]
2 years ago
6

If 25 kg of ice at 0C is combined with 4 kg of steam at 100C, what will be the final equilibrium temperature (in C) of the sy

stem? Latent heat of fusion of ice = 3.34 x 105 J.kg-1 Latent heat of steam = 2.23 x 106 J.kg-1 Specific heat of water = 4180 J.kg-1.K-1

Chemistry
2 answers:
givi [52]2 years ago
6 0

Try the suggested option; answer is marked with red colour (18.4953 °C).

All the details are in the attached picture.

Luden [163]2 years ago
5 0

Answer:

=18.5°C

Explanation:

Heat lost by steam=heat gained by ice.

ΔH(ice)=ΔH(steam)

ΔH=MC∅ where M is the mass of the substance, C is the specific heat capacity and ∅ is the temperature change of the substance.

Let final temperature be y°C

Then ΔH for ice=MLf+MC(y-0)°C

ΔH for steam= Mlv+MC(100-y)°C

(25kg×3.34×10⁵J/kg)+25kg×4180J/kgK×y°C= 4kg×2.23×10⁶J/kg+4kg×4180J/kgK×(100-y)

8350000J +104500y =8920000J+1672000J-16720y

Collect like terms together.

104500y+16720y=8920000J+1672000J-8350000J

121220y=2242000

y=18.5K

Final temperature= 0°C +18.5°C

=18.5°C

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