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Which material is a heat source for the other three materials?
Presented is a closed energy system containing equal masses of iron (55 degrees C), basalt (45 degrees C), and water (20 degrees C); air is 30 degrees C.
Which material is a heat source for the other three materials? 1) iron 2) basalt 3) air 4) water - Is the answer iron because it has the most heat?
During the first hour, the total energy in the system will 1) decrease, only 2) increase, only 3) decrease, then increase 4) remain the same - (Shouldn't it remain the same because of the conservation of energy?)
Also during the first hour, the temperature of the water will 1) decrease 2) increase 3) remain the same - (I think it will increase because it acts as a heat sink? Or maybe it will remain the same because water has the highest specific heat?)
Helping me understand one or all of these would be a great help!

What would be the temperature of the resultant mixture?
The question in my textbook reads
100 g of water is supercooled to -${10}^{\circ <!--\; \circ \; -->}$C. At this point, due to some disturbance mechanised or otherwise some of it suddenly freezes to ice. What will be the temperature of the resultant mixture and how much mass would freeze?
I am well aware of the concepts and principle of Calorimetry but I can't seem to get a clear picture of the question.
My doubts include -
1.Why doesn't the whole water in the beaker freeze?
2.Why does some part of the water in the beaker act as a source of heat to raise the temperature of the remaining water?
3.Why is the equilibrium attained at $0^{\circ <!--\; \circ \; -->}$C?

Is the loudness of a ball that you drop a measure of the amount of energy you put in to lift it?
When you lift a heavy object off the floor and raise it then you've exerted an amount of work/energy on that body. When you release the heavy object gravity exerts the same work/energy. But when you lift the ball nothing dramatic happens. You lift it and that's it. When you drop the ball there is a loud bang when the ball hits the floor. What accounts for this difference? It seems that it must be that there is an equivalent amount of energy in your body as in the sound (assuming that the floor doesn't change so that most of the energy goes into sound). Is this accurate reasoning? Can you do sound calorimetry?

How can scintillation gamma-spectrometers work given that track length is different for different angles?
As far as I understand, the basic principle of gamma spectrometer is simple - gamma ray hits scintillator, it generates number of photons which roughly proportional to gamma ray energy. Then we need to detect these secondary photons and plot a nice graph.
But as far as I see, scintillator size is typically too small to completely absorb gamma photon. Which also means that gamma photons coming from different angles will have different track lengths inside scintillator - and should generate different amount of light for the same gamma ray energy.
So, how can this work at all?

Calorimetry Problem
I was doing a problem in thermodynamics where the net heat is 0.
I don't understand why if you have say a copper calorimeter with water at say 15 °C and add a mass of copper at a higher temperature say 90 °C that when calculating the final temperature you would use for the copper piece this in the formula:
$Q=mc(T_f-<!--\; -\; -->T_i)$
$Q=mc(T_f-<!--\; -\; -->T_i)$
Where f is for final and i is for initial. Say mass is 0.3 kg. I was told that regarding $T_f-<!--\; -\; -->T_i$ I would need to use $90-<!--\; -\; -->T$. This problem was to work out the final temperature of the system. Why was the initial 90 °C of the added copper substituted with $T_f$ in the heat equation?

A Calorimetry Problem
I have a question in calorimetry from an old competitive exam. The question is:
The temperature of $100$ grams of water is to be raised from ${24}^{\circ <!--\; \circ \; -->}$C to ${90}^{\circ <!--\; \circ \; -->}$C by adding steam to it. Find the mass of steam required.
I tried attempting the question by assuming that the added steam would convert back to water and thus lose an amount of heat calculated by the latent heat of vaporization. Additionally, to cool from 100 degrees to 90 degrees, the 'watered steam' will contribute some additional heat. I equated this to the heat gained by the 100 grams of water to reach 90 degrees from 24 degrees. But I am not getting the right answer!
I would appreciate any help on this matter. Thank you :)

What quantity remains conserved in a calorimeter when two bodies at different temperatures are mixed?
This is the question:
Two identical bodies at different temperatures are mixed in a calorimeter. Which of the following quantities remains conserved?
1) Sum of the temperatures of the two bodies 2) Total heat of the two bodies 3) Total internal energy of the two bodies 4) Internal energy of each body
As we take a calorimeter to be an isolated system, internal energy of the two bodies would remain conserved. But why won't the total heat be conserved? The principle of calorimetry states that total heat gained is equal to total heat lost. I can't figure out what I'm missing here, and I'm very confused. Any help is appreciated.

What is MIP hypothesis?
While reading about muon analysis I read that finding calorimetry segments along with the tracks in the silicon tracker for muons helps us find a subset of tracks compatible with the MIP hypothesis. I was looking for what MIP means and all I could find is Minimum Ionizing Particle. Does MIP mean that and what is the MIP hypothesis?

Calorimetry - Emitted Joules
How can one calculate the total amount of emitted joules from an object with a temperature that isn't constant? A great start is this formula:
P = σ • A • (T^4)
If the formula is translated from sybmols to units it will look like this:
J/s = σ • (m^2) • (K^4)
(J/s) is joules emitted per second from an object. ( σ ) is the Stefan Boltzmann constant, 5.67•(10^−8). (m^2) is the area of the object. (K^4) is the temperature of the object, in Kelvin, to the power of 4. Now, if I transform the formula:
J = σ • (m^2) • (K^4) • s
Now one can get total amount of joules emitted ( J ) during a certain time ( s ), if one know the temperature and area. To the tricky part: what if the temperature isn't constant? The temperature will depend on how many joules that has already been emitted. And the joules that are being emitted will depend on the temperature.
How can I solve this? Do I need to combine this with an other formula?
Best regards! Please comment if something is unclear!

Final temperature of a mixture of water and ice in a copper container
Lets say I have ice of mass $m_i$ and initial temperature $T_i$ and specific heat $s_i$ . And I have water of mass $m_w$ and initial temperature $T_w$ and specific heat $s_w$ . I have put both this water and ice in a copper container with mass $m_{cu}$ , specific heat $s_{cu}$ and initial temperature $T_w$ . What will be the final temperature $T$ of the system ? ( The Latent heat of freezing of water is $L_f$ )
Here ,
$Q_1=m_i{s}_i(T-<!--\; -\; -->T_i)+m_i{L}_f$
$Q_2=m_w{s}_w(T-<!--\; -\; -->T_w)+m_w{L}_f+m_{cu}{s}_{cu}(T-<!--\; -\; -->T_w)$
According to Calorimetry ,
$Q_1=Q_2$
$m_i{s}_i(T-<!--\; -\; -->T_i)+m_i{L}_f=Q_2=m_w{s}_w(T-<!--\; -\; -->T_w)+m_w{L}_f+m_{cu}{s}_{cu}(T-<!--\; -\; -->T_w)$
I can use this equation to calculate $T$ , But then I ran into a problem. In this case , I don't know which is happening , the freezing of water , or the melting of ice , so i can't just put both $m_w{L}_f$ and $m_i{L}_f$ in the equation and call it done. I might have to remove one of them depending on which thing is not happening (freezing or melting) .To make matters worse , The water or ice may just freeze or melt partially. Is there another equation that can solve for the final temperature while also taking all of this into account?

Entropy change in a calorimetry problem
A standard textbook problem has us calculate the change in entropy in a system that undergoes some sort of heat exchange. For example, object $A$ has specific heat $c_a$ and initial temperature $T_A$ and object $B$ has specific heat $c_b$ with initial temperature $T_B$. They are they put in contact with each other until they reach thermal equilibrium, and our goal is to find the total entropy change of the system.
The standard solution is to use
$S=\int <!--\; \int \; -->\frac{dQ}{T}$
where $dQ=mcdT$. But the above integral is only satisfied for reversible processes, whereas this heat exchange is clearly irreversible.
The usual workaround for this is to pick some reversible path and calculate the entropy change on our "fake" path, since entropy is a state variable. For example, in the free expansion of an ideal gas, we pick calculate the entropy change along an isotherm that carries us along the expansion to find the true change in entropy.
My question is - what exactly is the reversible path we are using when we use $dQ=mcdT$?

Heat transfer vs calorimetry equation
I'm trying to understand the difference between the equations
$Q=mc\mathrm{\Delta <!--\; \Delta \; -->}T$
and,
$\frac{dQ}{dt}=\frac{-<!--\; -\; -->\mathrm{\kappa <!--\; \kappa \; -->}\mathrm{\alpha <!--\; \alpha \; -->}\mathrm{\Delta <!--\; \Delta \; -->}T}{l}$
Suppose, we have two metal rods at $T_1$ and $T_2$ temperature, and we connect them at the end. I want to know, what do the two above equations tell us, regarding this system.
My understanding is that the first equation tells us, that the two systems will reach an equilibrium temperature, and the heat gained by one rod would be the heat lost by the other. This final temperature would depend upon the relative masses, and heat capacities. However, the first equation tells us, how much heat flows out of the first rod, into the second rod, and the final temperature.
Where does this second equation come from, and what does it tell us ? It seems to me, that the second equation would give us the heat transferred from one rod to the other, in some given time, unlike the first one that gives us the total heat transferred.
Combining the two above equations, we can derive the partial differential equation, called the heat equation. My question is, what does my heat equation tell me ? The first equation tells me, the final temperature of two bodies in contact, provided I know the initial temperature, mass and heal capacities.
What does the heat/diffusion equation tell me then ? Does it tell me, the distribution of heat at some given time ?
Does it mean, if I use the heat equation for our problem with the two rods at different temperatures, I'd get the same final temperature as the $Q=mc\mathrm{\Delta <!--\; \Delta \; -->}T$ equation, after some time, and this temperature would be constant ?

Entropy change is defined as the amount of energy dispersed reversibly to or from the system at a specific temperature. Reversivility means that the temperature of the system must remain constant during the dispersal of energy to or from the system . But this criterion is only fulfilled during phase change & isothermal processes. But not all processes maintain constant temperature;temperature may change constantly during the dispersal of energy to or from the system . To measure entropy change ,say, from $300K$ & $310K$, the range is divided into infinitesimal ranges ,then entropy is measured in that ranges and then is integrated . But I cannot understand how they have measured entropy change in that infinitesimal ranges as there will always be difference between the temperature however small the range might be . What is the intuition behind it? Change of entropy is measured at constant temperature,so how can it be measured in a range ? I know it is done by definite integration but can't getting the proper intuition . Also ,if by using definite integration to measure change, continuous graph must be there(like to measure change in velocity,area under the graph of acceleration is measured) . So what is the graph whose area gives change in entropy? Plz help me explaining these two questions.

Water equivalent of calorimeter
I am trying to solve numerical problem based on calorimetry. The water equivalent of a calorimeter is given in grams. But how is it possible as
w= mc and unit of m*c is Cal/°C.
where w= water equivalent
m= mass of the water
c= specific heat of water

Equilibrium temperature of closed system
Body X of temperature 0° C is brought into thermal contact with body Y of temperature 100° C. X has specific heat capacity higher than of Y. The masses of X and Y are equal.
By my reasoning, the final equilibrium temperature should lie between 0° C and 50° C. Is this correct?
Edit: 1) The bodies are in thermal contact only with one another; they are in a closed system.
2) My reasoning:
$Q_x=m_x{c}_x\mathrm{\Delta <!--\; \Delta \; -->}{T}_x$
$Q_y=m_y{c}_y\mathrm{\Delta <!--\; \Delta \; -->}{T}_y$
$Q_x=Q_y$, $m_x=m_y$
$c_x\mathrm{\Delta <!--\; \Delta \; -->}{T}_x=c_y\mathrm{\Delta <!--\; \Delta \; -->}{T}_y$
If $c_x$ is higher than $x_y$, then $\mathrm{\Delta <!--\; \Delta \; -->}{T}_x$ must be lower thab $\mathrm{\Delta <!--\; \Delta \; -->}{T}_y$, so the equilibrium temperature must lie below 50° C.

How to derive the formula for heat produced due to electricity correctly from Joule's laws for heating?
From Joule's laws, we get this:
$H\propto <!--\; \propto \; -->I^{2}Rt$
$⟹<!--\; ⟹\; -->H=K{I}^{2}Rt...(i)$
Now, we have to find/define the value of K. According to my book, when $1A$ of current passes through a conductor of $1\mathrm{\Omega <!--\; \Omega \; -->}$ for $1s$, $1J$ heat is produced. If that's the case, then from $(i)$ we get this:
$1=K\times <!--\; \times \; -->1\times <!--\; \times \; -->1\times <!--\; \times \; -->1$
$⟹<!--\; ⟹\; -->K=1$
Therefore, we get our nice little formula:
$H=K{I}^{2}Rt$
My question with the above derivation is, how did we find that when $1A$ of current passes through a conductor of $1\mathrm{\Omega <!--\; \Omega \; -->}$ for $1s$, $1J$ heat is produced? Through experimentation? What was the name of the experiment and who conducted it?

Why the E of the time component 4-momentum is the total energy and not another?
The time component of the 4-momentum is $E/c$, and I saw that it is the "total energy" and from here we can derive the formula $E^2=(pc{)}^2+m^2{c}^4$
$E$ is the total energy? Can't it be some multiple of it or just some other energy?

What calorimetry approach is this?
Question:
Find the final tempertature of the system when $5gm$ of steam at ${100}^{\circ <!--\; \circ \; -->}C$ is mized with $31.5gm$ ice at $-<!--\; -\; -->{20}^{\circ <!--\; \circ \; -->}C$
Constants:
Specific heat of ice and specific heat of steam is $0.5cal/gm{/}^{\circ <!--\; \circ \; -->}C$
Specific heat of water is $1cal/gm{/}^{\circ <!--\; \circ \; -->}C$
Latent heat of fusion is $80cal/gm$
Latent heat of vaporization is $540cal/gm$
My attempt:
I usually approach these problems by setting the final temperature as $T$. Then, I equate heat lost by steam to heat gained by ice. I have been taught this method and I am adept at it.
My problem: The solution given in my workbook is:
Heat released when $1g$ of steam at ${100}^{\circ <!--\; \circ \; -->}C$ is cooled to $-<!--\; -\; -->{20}^{\circ <!--\; \circ \; -->}C$ is $730cal$. Now, this heat is given to be given to $1+6.3g=7.3g$ of ice. This raises the temperature by ${10}^{\circ <!--\; \circ \; -->}C$
This is a new method for me, and I am unable to understand it. I wish to know:
How does this method work? In what calorimetry situations can this method be applied?
NOTE: The final answer is ${10}^{\circ <!--\; \circ \; -->}C$