Meaning of ##\Delta T## in the context of calorimetry

[JC2000]

Homework Statement

A sphere of 0.047 kg aluminium is placed for sufficient time in a vessel containing boiling water, so that the sphere is at 100 °C. It is then immediately transfered to 0.14 kg copper calorimeter containing 0.25 kg water at 20 °C. The temperature of water rises and attains a steady state at 23 °C. Calculate the specific heat capacity of aluminium.

Relevant Equations

Mass of aluminium sphere $(m_1) = 0.047 kg$
Initial temperature of aluminium sphere$=100 °C$
Final temperature $= 23 °C$
Change in temperature $(∆T)=(100 °C-23°C)= 77 °C$
Let specific heat capacity of aluminium be $s_{Al}$.

The amount of heat lost by the aluminium sphere$= m_1s_{Al}∆T=0.047kg×s_Al ×77°C$
Mass of water $(m_2) = 0.25 kg$
Mass of calorimeter $(m_3) = 0.14 kg$
Initial temperature of water and calorimeter$=20 °C$
Final temperature of the mixture $= 23 °C$
Change in temperature $(∆T_2) = 23 °C – 20 °C = 3 °C$
Specific heat capacity of water $(s_w)= 4.18 × 103 J kg–1 K–1$
Specific heat capacity of copper calorimeter $= 0.386 × 103 J kg–1 K–1$

Heat gained by Calorimeter and Water = Heat Lost by Aluminium sphere

My question :

My understanding is that $\Delta T = T_f - T_i$

In the case of the calorimeter and water this is consistent with the solution. Yet for the sphere $\Delta T = T_i - T_f$?

 

[DrClaude]

$\Delta T = T_f - T_i$ always. For the aluminum sphere, $\Delta T < 0$, which is consistent with the fact that it loses heat, i.e., $\Delta E < 0$.

 

[JC2000]

So in the context of the problem do they mean $|\Delta T|$? Wouldn't a negative $RHS$ affect the sign of the solution?

 

[DrClaude]

Heat gained by Calorimeter and Water = Heat Lost by Aluminium sphere

This is where the sign comes in. You wrote "gained" on the LHS and "lost" on the RHS, which is equivalent to a change of sign.

 

[JC2000]

Really sorry if I am getting stuck on something trivial :
What I meant was : $m_2 s_w ∆T_2 + m_3s_{cal}∆T_2 = m_1s_{Al}\Delta T$
Now if RHS were to be negative then $s_{Al}$ would be negative(?)

 

[DrClaude]

What I meant was : $m_2 s_w ∆T_2 + m_3s_{cal}∆T_2 = m_1s_{Al}\Delta T$

That should be
$m_2 s_w ∆T_2 + m_3s_{cal}∆T_2 = - m_1s_{Al}\Delta T$
since it is
$$\Delta E_{\textrm{calorimeter} + \textrm{water}} = - \Delta E_{\textrm{aluminum}}$$

 

[JC2000]

Ah! So then on using $\Delta T$ correctly the signs, work out.
Seems my book expects this to be understood and moves on with both LHS and RHS having the same sign.
Thank you, I understand now!

 

[Chestermiller]

Another way of looking at this is that the total (internal) energy of the combined system U remains unchanged. So, $$\Delta U=m_{Al}s_{Al}(T_f-T_{Al,0})+m_{w}s_{w}(T_f-T_{w,0})+m_{Cu}s_{Cu}(T_f-T_{Cu,0})=0$$Now all the temperature changes are 'final' minus 'initial.'

 

[JC2000]

! Never thought of looking at that way!

Is this, so to say a 'newbie' problem (the inability to see something in multiple different ways?) which sort of goes away as I wade further into the subject, or should I be actively be doing something to gain this perspective? Thank you!

 

[Chestermiller]

! Never thought of looking at that way!

Is this, so to say a 'newbie' problem (the inability to see something in multiple different ways?) which sort of goes away as I wade further into the subject, or should I be actively be doing something to gain this perspective? Thank you!

Right now you have no experience and, moreover, the material has been presented to you in only one way. Later on, in thermodynamics, it will be presented to you in the way that I showed. With more experience and more training, you will be able to look at things in a variety of different ways.

 

[JC2000]

Thank you very much for your advice!

moreover, the material has been presented to you in only one way.

Do you mean in terms of statistical thermodynamics vs classical thermodynamics, etc or in terms of different textbooks explaining a topic 'differently'(for a beginner)?

With more experience and more training, you will be able to look at things in a variety of different ways.

I understand that there are a number of posts in extensive detail with regard to meta learning, but off the cuff are there some things I should be doing actively to help this process (other than reading/thinking about topics and solving diverse problems)?

 

[DrClaude]

I understand that there are a number of posts in extensive detail with regard to meta learning, but off the cuff are there some things I should be doing actively to help this process (other than reading/thinking about topics and solving diverse problems)?

The simplest is to read from as many different sources as possible. Different authors will present things in different ways, and sometimes a particular author will present a particular topic in a way that clicks with you. Even when this is not the case, diverse presentations can give you a feel for the subtleties of a subject, even when you don't yet master the theory behind it.

 

[Chestermiller]

Thank you very much for your advice!
Do you mean in terms of statistical thermodynamics vs classical thermodynamics, etc or in terms of different textbooks explaining a topic 'differently'(for a beginner)?

The latter.

I understand that there are a number of posts in extensive detail with regard to meta learning, but off the cuff are there some things I should be doing actively to help this process (other than reading/thinking about topics and solving diverse problems)?

Nothing replaces solving lots of diverse problems.