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

• JC2000
In summary: But in addition to that, it can be helpful to try to explain what you have learned to others. This can help you identify any gaps or misunderstandings in your own understanding, and also force you to think about things from different perspectives. You can also try teaching the material to yourself in different ways - for example, if you usually read the textbook, try watching a lecture on the topic, or vice versa. Experimenting with different study strategies can also help you develop a deeper understanding of the material.

#### 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##?

##\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?

JC2000 said:
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(?)

JC2000 said:
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!

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.'

DrClaude and 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!

JC2000 said:
! 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

Chestermiller said:
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)?

Chestermiller said:
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)?

JC2000 said:
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 and JC2000
JC2000 said:
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.

JC2000 and DrClaude

## 1. What does ##\Delta T## represent in calorimetry?

##\Delta T## represents the change in temperature of a substance in a calorimetry experiment. It is calculated by subtracting the initial temperature from the final temperature.

## 2. Why is ##\Delta T## important in calorimetry?

##\Delta T## is important because it allows us to measure the amount of heat energy exchanged during a chemical reaction or physical process. This information is crucial in understanding the properties of a substance and its reaction with other substances.

## 3. How is ##\Delta T## measured in a calorimetry experiment?

##\Delta T## is measured using a thermometer that is placed in the substance being studied. The initial and final temperatures are recorded and then the change in temperature is calculated.

## 4. What is the significance of a negative ##\Delta T## in calorimetry?

A negative ##\Delta T## indicates that the substance has lost heat energy, meaning that the reaction or process is exothermic. This information is important in determining the energy released by the substance and its reactivity.

## 5. How does the specific heat capacity affect ##\Delta T## in calorimetry?

The specific heat capacity of a substance determines how much heat energy is required to raise its temperature by 1 degree Celsius. Therefore, a substance with a higher specific heat capacity will have a smaller ##\Delta T## compared to a substance with a lower specific heat capacity, for the same amount of heat energy exchanged.