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

AI Thread Summary
To calculate the final mass of water in a calorimetry experiment, one must equate the heat lost by steam to the heat gained by the system components. The specific heat and temperature changes for each material must be correctly applied, ensuring that different temperature changes are not confused. The equation for the mass of steam must include proper parentheses to clarify the calculation of heat transfer. The temperature change for steam is indeed -80°C, but should be used as +80°C when calculating the heat supplied. Proper notation and attention to detail in the equations are crucial for an accurate solution.
unknown217
Messages
2
Reaction score
0

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)?
 
Physics news on Phys.org
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
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top