Specific heat and possibly latent heat too..

AI Thread Summary
The discussion revolves around a physics homework problem involving the mixing of ice and steam to determine the final temperature. The initial equations used by the student included both latent heat and specific heat, but they struggled with the calculations. After realizing a mistake in their equation setup, they received guidance on correcting the temperature change terms. The conversation highlights the importance of accurate calculations and proper equation formulation in thermodynamic problems. Ultimately, the student expresses gratitude for the assistance in resolving their confusion.
shannon.leigh
Messages
9
Reaction score
0

Homework Statement


0.5kg of ice at 0 degrees celcius is mixed with 0.1kg of steam at 100 degrees celcius. what will be the final temperature?


Homework Equations


delta Q= m*c*delta T

and at first i thought delta Q= m*Lf ,but then i changed my mind. . .

The Attempt at a Solution



In my first attempt, i went

Lf*m(of ice) + m*c*(Tf-Ti)(of the now liquid water)=Lv*m(of steam)+ c*m*(Tf-Ti)(of now liquid water)

so 3.34*10^5*0.5 +0.5*4200*(T-0)=2.26*10^6*0.1+4200*0.1*(100-T)
167000+2100T=226000+42000-420T
2520T=435000
T=435000/2520
=172.62
Which is wrong!

so then i got rid of the latent heat and fusion, because i thought they were irrelevant to what i was trying to figure out and went
2100T=42000-420T
T=42000/2520
=16.67
Which is still wrong!

HELP ME PLEASE!

thankyou (:
 
Physics news on Phys.org
shannon.leigh said:
167000+2100T=226000+42000-420T
2520T=435000

Do the simplification again, the RHS is not 435000.

ehild
 
Are you serious? I just screwed up the calculations?
Damn my terrible skills with a calculator!
And thankyou thankyou thankyou for pointing that out ehild!
I am forever in your debt. :)
 
To tell the truth, there was a mistake in your equation, but you proceeded all right till the last row.

"Lf*m(of ice) + m*c*(Tf-Ti)(of the now liquid water)=Lv*m(of steam)+ c*m*(Tf-Ti)(of now liquid water)"

The red text should be Ti-Tf.

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 'Calculation of Tensile Forces in Piston-Type Water-Lifting Devices at Elevated Locations'
Figure 1 Overall Structure Diagram Figure 2: Top view of the piston when it is cylindrical A circular opening is created at a height of 5 meters above the water surface. Inside this opening is a sleeve-type piston with a cross-sectional area of 1 square meter. The piston is pulled to the right at a constant speed. The pulling force is(Figure 2): F = ρshg = 1000 × 1 × 5 × 10 = 50,000 N. Figure 3: Modifying the structure to incorporate a fixed internal piston When I modify the piston...
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