Double-Check Homework: Specific Heat & Melting Ice

AI Thread Summary
The discussion focuses on verifying calculations related to specific heat and melting ice. For the first problem, the specific heat of the unknown metal block is calculated to be approximately 0.0904 cal/g°C. In the second problem, it is determined that 42.19 g of ice melts when 45 g of water at 75°C is added, leaving some ice remaining. The final temperature in the cup remains at 0°C since not all the ice has melted. Overall, the calculations appear to be correct.
veronicak5678
Messages
143
Reaction score
0
Hoping someone could double-check for me...

Homework Statement



1) A 90 g block of unknown metal is heated to 86 degrees C and put in an insulated cup with 175 g water at 23 degrees C. At equilibrium, they are both 25.8 degrees C. What is the specific heat of the metal block?

2) A well insulated cup contains 145 g of ice at 0 degrees C. We pour in 45 g watwer at 75 degrees C.
a- Once equilibrium is established, how many grams of ice remain?
b- What is the final temp in the cup?




The Attempt at a Solution


1) c ( 90 g) (-60.2 degrees c) + 1 cal/g degrees C) (175 g) (2.8degrees C) = 0
c = 0.0904 cal/g degree C

2)
a- energy available to melt ice: 1 cal/ g Celcius (45 g) ( 75 degrees ) = 3375 cal
energy needed to melt ice: 145 g (80 cal/g) = 11600 cal
not all will melt

80 cal /g (mass melted) = 3375 cal
mass melted = 42.19 g

b- Temp is 0 since there is still ice.
 
Physics news on Phys.org
Looks good to me!
 
OK, great. Thanks for looking it over!
 
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