How Do You Calculate the Final Mass of Water in a Calorimetry Experiment?

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Homework Help Overview

The problem involves a calorimetry experiment where a copper container holds water and ice, and steam is introduced to determine the final mass of water after thermal equilibrium is reached. The specific heat capacities and mass of the substances involved are provided, and the system is assumed to be insulated.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the heat transfer equations, questioning the correct application of temperature changes for steam and the other components. There is a focus on ensuring proper notation and parentheses in the equations used.

Discussion Status

Some participants have offered guidance on the correct formulation of the equations, emphasizing the need for clarity in the use of symbols and parentheses. Multiple interpretations of the equations are being explored, particularly regarding the temperature changes and their implications for the calculations.

Contextual Notes

Participants highlight the importance of distinguishing between different temperature changes in the system and the potential for confusion in notation. The discussion reflects an ongoing effort to clarify these aspects without reaching a final conclusion.

unknown217
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Homework Statement


A 0.080kg copper container (specific heat: 387Jkg-1K-1) contains 0.30kg of water and 0.040kg of ice at 0°C. Steam at 100°C is passed into the water and its temperature stabilizes at 20.0°C. Find the mass of the water left in the container assuming the system is insulated.

Homework Equations


Q= mcΔT
Q= mL

The Attempt at a Solution


Q lost = Q gained
msteamcΔT + msteamL = mCucΔT + mH20cΔT + micecΔT + miceL
msteam = mCucΔT + mH20cΔT + micecΔT + miceL / cΔTsteam + Lsteam
Total mass water= mass steam + water + ice

Is this correct?
Is ΔT for steam -80 (20-100°C)?
 
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unknown217 said:

Homework Statement


A 0.080kg copper container (specific heat: 387Jkg-1K-1) contains 0.30kg of water and 0.040kg of ice at 0°C. Steam at 100°C is passed into the water and its temperature stabilizes at 20.0°C. Find the mass of the water left in the container assuming the system is insulated.

Homework Equations


Q= mcΔT
Q= mL

The Attempt at a Solution


Q lost = Q gained
msteamcΔT + msteamL = mCucΔT + mH20cΔT + micecΔT + miceL
msteam = mCucΔT + mH20cΔT + micecΔT + miceL / cΔTsteam + Lsteam
Total mass water= mass steam + water + ice

Is this correct?
Is ΔT for steam -80 (20-100°C)?

Welcome to PF!

Do not use the same symbol ΔT for different things. The change of temperature of water is not the same as the change of temperature of the deposited steam.
Your equation means that the heat supplied by the steam is equal to the heat absorbed by the colder parts of the system.
It is right that the change of temperature is -80° for the deposited steam, but you have to use 80° when you calculate the supplied heat.
The equation for msteam is not correct without parentheses around cΔTsteam + Lsteam

ehild
 
msteam = mCucΔTCu + mH20cΔTH20 + micecΔTice + miceL / (cΔTsteam + Lsteam)

Is this right?
 
You need parentheses also around the nominator. And be sure using correct sign in the denominator.

ehild
 

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