Temperature Fusion (heat gain = heat loss)

[aodhowain]

Homework Statement

If 10 kilogram of ice at zero degrees celsius is added to 20 kilogram of steam at 100 degrees celsius, what is the temperature of the resulting mixture?

Homework Equations

uhm... perhaps how many solutions are required to arrive at the answer?

The Attempt at a Solution

I and some of my classmates arrived at 16.11 degrees celsius but my instructor said its wrong. He said the answer is around 40 degrees celsius. Now, he gave it as an assignment to us to find the correct solution.

 

[Borek]

In no way that's advanced physics.

Show how you got 16 deg answer. What equations have you used?

 

[aodhowain]

What equations have you used?

Q_E = Q_A
Heat Evolve = Heat Absorb
M_EC_Edelta T_E = M_AC_Adelta T_A

 

[Borek]

Q_E = Q_A
Heat Evolve = Heat Absorb

So far so good.

M_EC_Edelta T_E = M_AC_Adelta T_A

That's not all.

I have asked you to show how you got the wrong answer.

 

[aodhowain]

M_EC_Edelta T_E = M_AC_Adelta T_A
(10 kg)(0.50 cal/kg*C)(tf - 0C) = (2 kg)(0.48 cal/kg*C)(100C - tf)
5 cal/C * tf = 96 cal - 0.96 cal/C * tf
5.96 cal/C * tf = 96 cal
tf = (96 cal)/(0.96 cal/C)
tf = 16.11 C

 

[Borek]

As I told you - you are missing part of the equation. Two parts, to be precise. What you did would be OK for a mixture of water at 0 deg C and water at 100 deg C - but you start with ice and steam.

 

[fluidistic]

As I told you - you are missing part of the equation. Two parts, to be precise. What you did would be OK for a mixture of water at 0 deg C and water at 100 deg C - but you start with ice and steam.

I agree. A bit more help, heat from hot water has to be transferred to the ice to melt it, even though the temperature does not change. Quantify this heat.

 

[aodhowain]

I think I got the solution to arrive at 40 degrees celsius. Please verify if my answer is correct.

Q out of steam = Q into ice (for c I use kcal/kgC but You can just as easily use J/kgC)

Q out = m*Lv + m*c*deltaT = m*(Lv + c*deltaT) = 2.0kg*(540kcal/kg + 1.0kcal/kg/C*(100-Tf))

Q in = m*Lf +m*c*deltaT = m*(Lf =c*deltaT) = 10kg*(80kcal/kg + 1.0kcal/kgC*(Tf -0))

Setting these equal gives 2.0*540 +2.0*100 -2.0*Tf = 10*80 + 10*Tf

Solving (10+2)*Tf = 2*540 +2*100-10*80 = 480 so Tf = 480/12 = 40degC

 

[Borek]

I have just skimmed, looks reasonably.

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