Calorimetrey Problem - Final Temperature of water

AI Thread Summary
In the calorimetry problem, 0.50 kg of water at 80°C is mixed with 0.050 kg of ice at -5.0°C in an insulated container. The goal is to find the final temperature of the resulting water after all the ice melts. The calculations involve using the heat transfer equations, accounting for the specific heat of water and ice, as well as the latent heat of fusion. The initial attempt resulted in a final temperature of 68°C, but the correct answer is 65°C, indicating a potential error in sign usage for heat transfer. The solution requires careful consideration of the heat gained by the ice and the heat lost by the water.
forestmine
Messages
201
Reaction score
0
Calorimetrey Problem -- Final Temperature of water

Homework Statement



In an insulated container, 0.50kg of water at 80C is mixed with 0.050kg of ice at -5.0C. After a while, all the ice melts, leaving only the water. Find the final temperature T_f of the water. The freezing point of water is 0 C.

Homework Equations


Q=mcΔT
Q=mL (heat transfer in a phase change)
specific heat of water c = 4186 J/kgK
specific heat of ice c = 2100 J/kgK
latent heat of fusion for water = 3.34 x 10^5 J/kg

The Attempt at a Solution



Here's my attempt at a solution which came to 68 degrees C (the correct answer is 65C). I've checked my math, so I don't think it was a simple computational error. I think I'm making a mistake in terms of my signs perhaps when it comes to heat entering and exiting.

Anyway, here's what I did.

Q(water) + Q(ice) + m(ice)L = 0

(.5)(4186)(T-80) + (.05)(2100)(T+5) + (.05)(3.34x10^5)=0

I understand that in the case of Q(water), heat is exiting, and in the case of Q(ice) heat is entering. And since the ice melts, in the case of mL, heat is also entering. I guess I'm just not getting something crucial, here.

Any help would be greatly appreciated! Thank you.
 
Physics news on Phys.org


The melted ice is water. Determine the heat needed to warm up the ice to the melting point, then calculate with the specific heat of water.

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