Heat required to heat water and ice

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SUMMARY

The discussion centers on calculating the heat required to raise the temperature of a system containing 2.40 kg of water and 0.450 kg of ice from 0.0°C to 28.0°C by condensing steam. The correct formula for the heat required is Qreq = miLf + (mi + mw)cΔT, leading to the calculation of steam mass as ms = -(miLf + (mi + mw)cΔT)/Lv. The final calculation yields 215 grams of steam, contradicting the physics book's stated answer of 190 grams, which neglects the mass of ice in the temperature change calculation. The user ultimately confirms their solution as correct after clarification.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically heat transfer.
  • Familiarity with specific heat capacity (c) and latent heat (Lf, Lv).
  • Ability to apply the equations Q = mcΔT and Q = ±mL in problem-solving.
  • Knowledge of phase changes and their associated energy requirements.
NEXT STEPS
  • Review the principles of heat transfer in thermodynamics.
  • Study the calculations involving latent heat and specific heat in different states of matter.
  • Practice similar problems involving phase changes and temperature changes in closed systems.
  • Explore the implications of thermal insulation in heat transfer scenarios.
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Students studying thermodynamics, physics educators, and anyone interested in understanding heat transfer calculations in systems involving phase changes.

Elias Waranoi
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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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