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tdusffx
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One kg of ice at 0°C is added to one kg of boiling water. The mixture comes to
equilibrium. What is the change in entropy in cal/K of the system? (Lf = 80cal/g)
a. 300
b. 100
c. 200
d. 50
e. 25
Qcold = -Qhot
Mw*Cw*dT + Lf*Mw = -Mw*Cw*dT + Mw*Lv
All has Mw in common, therefore it should cancel out
Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv
(1cal/g)(Tf - 0) + 80cal/g = -(1cal/g)(Tf-100) + 540cal/g;
I converted the latent heat of vaporation (2.26x10^6J/kg)*(1/4186cal*kg/J*g) ~
540cal/g
after equating the two sides, I got 280C for Tf
My answer for Tf doesn't seem reasonable because the Final Temp should be in between
0-100C
Anyways, My other problem is getting the dQ
I have no idea how to get it...or maybe I do..when it doubt I just ask people out!
dS = dQ/T
if my T is right then just finding dQ problem
Thanks in advance folks.
equilibrium. What is the change in entropy in cal/K of the system? (Lf = 80cal/g)
a. 300
b. 100
c. 200
d. 50
e. 25
Qcold = -Qhot
Mw*Cw*dT + Lf*Mw = -Mw*Cw*dT + Mw*Lv
All has Mw in common, therefore it should cancel out
Cw*dT + Lf*Mw = -Cw*dT + Mw*Lv
(1cal/g)(Tf - 0) + 80cal/g = -(1cal/g)(Tf-100) + 540cal/g;
I converted the latent heat of vaporation (2.26x10^6J/kg)*(1/4186cal*kg/J*g) ~
540cal/g
after equating the two sides, I got 280C for Tf
My answer for Tf doesn't seem reasonable because the Final Temp should be in between
0-100C
Anyways, My other problem is getting the dQ
I have no idea how to get it...or maybe I do..when it doubt I just ask people out!
dS = dQ/T
if my T is right then just finding dQ problem
Thanks in advance folks.