How much steam to raise vessel temperature from 0 to 28 C?

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SUMMARY

The discussion centers on calculating the amount of steam required to raise the temperature of a thermally insulated vessel containing 2.40 kg of water and 0.450 kg of ice from 0°C to 28°C. The user initially calculated 214.8 g of steam needed, while the book states the answer is 190 g. The key equations used include Q=mc∆t for heat transfer and Q=mL for phase changes. A critical oversight in the user's calculation was neglecting the cooling of the condensed steam from 100°C to 28°C, which significantly affects the total heat transfer required.

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  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with phase changes of water, including heat of fusion and heat of vaporization.
  • Proficiency in using the specific heat capacity formula.
  • Knowledge of conservation of energy in closed systems.
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  • Review the concept of heat transfer in thermally insulated systems.
  • Study the calculations involving heat of fusion and heat of vaporization for water.
  • Learn how to apply the specific heat capacity formula in multi-phase systems.
  • Explore the effects of temperature changes on phase changes in thermodynamics.
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Students studying thermodynamics, educators teaching heat transfer concepts, and anyone involved in calorimetry or energy conservation calculations.

JustinLiang
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Homework Statement



A vessel whose walls are thermally insulated contains 2.40kg of water and 0.450kg of ice, all at a temperature of 0.0 C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0 C? you can ignore the heat transferred to the container.

Homework Equations



Q=mc∆t
Q=mL
Water heat of fusion - 334000
Water heat of vaporization - 2256000
specific heat of water - 4190

The Attempt at a Solution



I am pretty sure my logic is right but my answer does not match with the book answer. I think the book may be wrong... The answer in the book is 190g.

What I did was:
Q(melting ice) + Q(water + melted ice from 0 C to 28 C) = Q(vaporization)
(0.45kg)(334000) + (0.45kg+2.4kg)(4190)(28K) = m(2256000)

I solved for m and I get 214.8g.

Am I doing anything wrong? If so what?

Thanks.
 
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JustinLiang said:

Homework Statement



A vessel whose walls are thermally insulated contains 2.40kg of water and 0.450kg of ice, all at a temperature of 0.0 C. The outlet of a tube leading from a boiler in which water is boiling at atmospheric pressure is inserted into the water. How many grams of steam must condense inside the vessel (also at atmospheric pressure) to raise the temperature of the system to 28.0 C? you can ignore the heat transferred to the container.

Homework Equations



Q=mc∆t
Q=mL
Water heat of fusion - 334000
Water heat of vaporization - 2256000
specific heat of water - 4190

The Attempt at a Solution



I am pretty sure my logic is right but my answer does not match with the book answer. I think the book may be wrong... The answer in the book is 190g.

What I did was:
Q(melting ice) + Q(water + melted ice from 0 C to 28 C) = Q(vaporization)
(0.45kg)(334000) + (0.45kg+2.4kg)(4190)(28K) = m(2256000)

I solved for m and I get 214.8g.

Am I doing anything wrong? If so what?

Thanks.

Don't forget that once the steam has condensed from steam at 100 C to water at 100 C, then that water at 100 C has to cool to 28 C.

I don't see anything changing temperature by 72 C in your calculation.
 

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