If 50.0 g of ice at 0 C is added to 150.0 g of water at 80.0 C?
What will the temperature of the mixture be when it reaches thermal equilibrium?
- 7 years agoFavorite Answer
mass of ice (m1) = 50 grams
mass of water (m2) = 150 grams
Temperature of water (T2)= 80.0 C
Temp. of ice (T1) = 0 C
Temperature of mixture (Tm)= ?
heat of melting ice(L) = 80 cal/g
heat capacity of water (c) = 1 cal/g
m1.L + m1.c.ΔTice = m2.c.ΔTwater
m1.L + m1.c.(Tm-T1) = m2.c.(T2-Tm)
50.80 + 50.1(Tm-0) = 150.1.(80-Tm)
4000 + 50Tm = 12000 -150Tm
200 Tm = 8000
Tm = 40 C
ok? rate please
- 7 years ago
I cannot do the math. I encountered a roadblock in my math that I cannot overcome. However, I can give you and any future people answering this question this tip:
You will need to find the total amount of energy transferred and include the heat of fusion of ice. Here is what I can help with.
The amount of energy transferred into the ice to melt it: Q=m*Hf 16700 J = .050*(3.34*10^5)
After considering the melted ice, we can now find the final temperature by using this formula:
The water we gained from melting the ice will be represented by "i", and temperature will be measured in Kelvins
Final Temperature = ((mass of i)*Ci*(initial temp)+(mass of w)*Cw*(initial temp)/((mass of i)*Ci+(mass of w)*Cw)
Final Temperature = (.050*4180*273+.150*4180*(initial temp))/(.050*4180+.150*4180)
Final Temperature = (57057+627*(initial temp))/(836)
The only variable I am missing is the temperature of the .150 kg of water after melting the ice.Source(s): Physics class - Thermodynamics
- 7 years ago
(50.0 * 0 + 150.0 * 80)/(50+150) = 60 C
This is assuming its a perfect system.