- #1
Ethan_Tab
- 57
- 2
Homework Statement
Hey folks, anyone who can help me with this one get a massive shoutout. Here goes: 75 kg of water at 45 degrees celsius is "mixed" with 70 kg of Aluminum at 2600 Degrees Celsius. What is the final state of both system and how much, If any, of liquid water will be left
Assume the following conditions.
Specific heat capacity :
Water--4.18X10^3
Aluminium-- 9.2X10^2
Latent Heat of fusion
Aluminium-- 6.6X10^5
Latent Heat of Vaporization
Water--2.3X10^6
Boiling points
Water-100
Melting points
Alumium--2519
Homework Equations
Q=mc(delta temperture)
L_fusion or vaporavation=mL_f or mL_v
I will represent the following as these variables
Ti= temperture intial
F=Final temperature of system
w=Water
a=alumium
The Attempt at a Solution
m_w*c_w(100-Ti_w)+m_w*Lv_w+m_w*c_w(F-100)+m_a*c_a(2519-Ti_a)+m_a*Lf_a+m_a*c_a(F-2519)=0
Factoring out the F out of the two terms and isolating for it I get:
F= [-m_w*c_w(55)-m_w*Lv_w+m_w*c_w(100)+m_a*c_a(81)-m_a*Lf_a+m_a*c_a(2519)]/(m_w*c_w+m_a*c_a)
This results in a negative number. Any Ideas where I went wrong?