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Calorimetry specific heat

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1. 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.


2. Homework Equations
Q= mcΔT
Q= mL


3. 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)?
 

ehild

Homework Helper
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1. 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.


2. Homework Equations
Q= mcΔT
Q= mL


3. 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?
 

ehild

Homework Helper
15,354
1,759
You need parentheses also around the nominator. And be sure using correct sign in the denominator.

ehild
 

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