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

AI Thread Summary
To raise the temperature of a vessel containing 2.40 kg of water and 0.450 kg of ice from 0 to 28 C, the heat required includes the melting of ice and heating the resulting water. The user calculated the necessary steam mass using the equations for heat transfer but arrived at 214.8 g, which does not match the book's answer of 190 g. A key oversight in the calculation was not accounting for the cooling of the condensed steam from 100 C to 28 C, which requires additional heat loss. This missing factor significantly impacts the total heat balance needed for the temperature increase. The discussion highlights the importance of including all temperature changes in thermal 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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