Why am I calculating the unknown mass of ice incorrectly?

[gabe99]

Homework Statement

1.5 dl of water at skin temperature of 36 °C was placed in a well-insulated thermos container. An icy piece of iron, including its mass, was added, weighing 230 g and at a temperature of 0 °C. After thermodynamic equilibrium was established, the ice had melted from the iron piece and the system's temperature was 11 °C. How many grams of ice were attached to the iron piece? The heat capacity of the thermos container can be assumed to be negligible, and the density of water can be estimated with three significant figures as 1000 kg/m³.

Relevant Equations

c_r = 0,449 kJ/kg*K (r is for Iron in my language)
c_v = 4,19 kJ/kg*K (v is for water)
s = 333 kJ/kg

The answer should be around 39g. I keep getting around 45g.

This is the calculation I tried

Any feedback on what I am doing wrong ?

Regards!

 

[Chestermiller]

Can you please add some text to describe what you are doing in your equation. I'm not able to figure out how you approached this.

 

[gabe99]

In my approach, I tried to emulate my equation based on what I learned in the theory of this topic:

"Heat is conserved in any such process, consistent with the law of conservation of energy. The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object."

In this case water, the warmer object lost heat, and the iron and ice gained heat

 

[Chestermiller]

Change in internal energy of original water = (100)(4.19)(-25)
Change in internal energy of ice = m(333)+m(4.19)(11)
Change in internal energy of iron = (230-m)((0.449)(11)
where all masses are in grams.

 

[kuruman]

"Heat is conserved in any such process, consistent with the law of conservation of energy. The amount of heat lost by a warmer object equals the amount of heat gained by a cooler object."

How do you express that with an equation?
You can see that your very first equation

is incorrect. The left hand side of the equation is a negative umber while the right hand side is positive. @Chestermiller provided you with expressions for the changes. How can you put them together in an equation?

 

[gabe99]

Would I put a "-" sign before the c_vm_v(-25) ? like -ΔQ_1=ΔQ_2 ?

As for the mass of water, should it no be 0.15 kg (since it is 1,5 dl) ?

 

[Chestermiller]

Would I put a "-" sign before the c_vm_v(-25) ? like -ΔQ_1=ΔQ_2 ?

The sum of the internal energy changes should be zero

As for the mass of water, should it no be 0.15 kg (since it is 1,5 dl) ?

Yes. I mistook it for 1.0 dl. If should be 150 gm in my equation.

 

[kuruman]

Would I put a "-" sign before the c_vm_v(-25) ? like -ΔQ_1=ΔQ_2 ?

That would make it positive. However, you have to understand why to understand what that means physically beyond what makes the algebraic equation correct.

If you write -ΔQ_1=ΔQ_2 it follows that ΔQ_2 + ΔQ_1 = 0. Explain what that equation is saying in plain English and without symbols.

As for the mass of water, should it no be 0.15 kg (since it is 1,5 dl) ?

You cannot mix decimal conventions. It would be 0.15 kg which would be the mass of 1.5 dl.

 

[gabe99]

ΔQ_2 + ΔQ_1 = 0. In a closed system the energy is conserved, no energy is lost or gained.

-ΔQ_1=ΔQ_2 . The amount of heat/energy lost by the water in this case is equal to the heat gained by the melting of the ice into water, the warming of the water, the warming of the iron.

Then I end up back at my equation again.

Edit: I forgot to put the minus sign in the beginning again, sorry.

 

[gabe99]

After the equation I tried solving for m_ice

 

[Chestermiller]

After the equation I tried solving for m_ice

So what did you get?

 

[gabe99]

So what did you get?

Rounded to 45 grams. The exercise says the answer should be 39 grams

 

[Chestermiller]

I get 39 C from my equation, replacing the 100 gm with 150 gm in post #4.

 

[Chestermiller]

ΔQ_2 + ΔQ_1 = 0. In a closed system the energy is conserved, no energy is lost or gained.

-ΔQ_1=ΔQ_2 . The amount of heat/energy lost by the water in this case is equal to the heat gained by the melting of the ice into water, the warming of the water, the warming of the iron.

Then I end up back at my equation again.

View attachment 353042

Edit: I forgot to put the minus sign in the beginning again, sorry.

The sum of the left side and the right side should be zero.

 

[gabe99]

Maybe a mistake in the exercise ?

 

[Chestermiller]

Maybe a mistake in the exercise ?

The exercise is correct. Maybe you need to brush up on your algebra for solving a linear algebraic equation in one unknown?

 

[gabe99]

Could be, no doubt. I just wondered since you said you changed the mass to 100g, and then got the desired result of 39g. The mass of the ice should be around 39 grams. I will check my algebra and come back

 

[Chestermiller]

Could be, no doubt. I just wondered since you said you changed the mass to 100g, and then got the desired result of 39g. The mass of the ice should be around 39 grams. I will check my algebra and come back

The initial mass of the water is 150 gm (1.5 dl). I mistakenly took it to be 1.0 dl. With 100 g of water initially, I would have obtained 24 gm for the final result.

Regarding you suggestion to use statistical mechanics to study the stability and possible convection, I would say that is an inappropriate choice. Anyone with any kind of experience would study these using viscous fluid mechanics, treating the gas as a continuum.

 

[gabe99]

That's the feedback that I'd like to receive. This is the approach we were taught.

I got it correct now, thank you all for the help.

 

[gabe99]

The mistake was the missing "-" sign in the left part like @kuruman pointed out, the algebra itself was correct.

 

[Chestermiller]

The mistake was the missing "-" sign in the left part like @kuruman pointed out, the algebra itself was correct.

Then if the minus sign were missing, you should have gotten -39 C.

 

[gabe99]

Then if the minus sign were missing, you should have gotten -39 C.

 

[kuruman]

Then if the minus sign were missing, you should have gotten -39 C.

That would be the case if the unknown in the equation

were $m_v$. The unknown mass of ice here is denoted by $m_j$.