Heat required to heat water and ice

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The discussion revolves around a physics problem involving the heat required to raise the temperature of a system containing water and ice. The user calculates the mass of steam needed to condense in order to achieve a temperature increase from 0.0°C to 28.0°C, arriving at a result of 215 grams. However, the physics book states the answer is 190 grams, leading to confusion about whether the user's calculations are correct. After further consideration, the user realizes the error was in not accounting for the ice in the heat capacity calculation. Ultimately, the user confirms their understanding of the problem and acknowledges the resolution.
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Is the answer in my physics book wrong or did I miss something?

Homework Statement


A vessel whose walls are thermally insulated contains 2.40 kg of water and 0.450 kg of ice, all at 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.

mw = 2.4 kg
mi = 0.45 kg
ΔT = 28 C°
c = 4190
Lf 334 * 10^3 J/kg
Lv 2256 * 10^3 J/kg

Homework Equations


Q = mcΔT
Q = ±mL

The Attempt at a Solution


Heat required to melt ice and warm water to 28 C°:
Qreq = miLf + (mi + mw)cΔT

Condensing steam of amount Qreq = -msLv
ms = -(miLf + (mi + mw)cΔT)/Lv
ms = -215 grams

My physics book tells me the answer is 190 grams, which I get if I don't account for mi in (mi + mw)cΔT. This seems wrong to me but I wanted to ask and make sure, every other time I've claimed that I found an error in my physics book I have been incorrect.
 
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When the steam condenses, is it at the final temperature?
 
Oops, thank you I got it right now.
 
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