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Specific heat is the amount of thermal energy required to increase/ decrease the temperature of a unit mass of a substance by 1K

This definition makes sense to me and is straightforward, however the sources I'm referring to say that specific heat is also a measure of a substance's storage ability. The term specific heat is often used in conjunction with heat storage. I'm not able to comprehend how is specific heat related to heat storage.

Edit: The answers have somewhat made sense to me that if we have two materials A and B at the same temperature initially having same mass then if A has a higher specific heat than B, then more energy will need to be given to A than B to result in same rise in temperature, so A has stored more energy in that temperature rise. This storage thing makes perfect sense to me when the body is just gaining energy but doesn't make sense when its gaining and losing at the same time (like from one end its gaining energy and losing from the other).

I'll try to explain my confusion with an example - Consider again the materials A and B (A has a higher specific heat than B). I make two blocks, completely identical, out of these materials A and B, and initially at the same temperature. Let us say lateral surfaces are insulated so that the heat transfer is 1D. I transfer 10kJ of energy to both the blocks from one side, since block A has a higher specific heat than B, will it store more energy than B? As in, block A stores 8kJ and leaves 2kJ whereas block B stores 2kJ and leaves 8KJ (arbitrarily assumed values)

If yes, how we can conclude that? We have stated that material A stores more energy than B, for same temperature rise, when only heat transfer is taking place to the body, how can I apply that concept to a case where heat transfer is taking place to and from the body simultaneously?

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  • $\begingroup$ Are you familiar with the concept of internal energy U? $\endgroup$ Commented Nov 2, 2021 at 19:50
  • $\begingroup$ Why do you have heat leaving A and B? $\endgroup$ Commented Nov 4, 2021 at 16:27
  • $\begingroup$ A heat sink at a lower temperature (than the initial common temperature of the bodies) is in contact with the bodies ,say, at the right. and similarly a heat source at a higher temperature(than the initial common temperature of bodies) is in contact with the bodies on the left. same heat source and same heat sink for both the bodies. $\endgroup$ Commented Nov 4, 2021 at 16:33
  • $\begingroup$ Are you asking what happens as a function of time (transient behavior), or are you asking what the temperature profiles are at final steady state? $\endgroup$ Commented Nov 4, 2021 at 18:20
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    $\begingroup$ @ChetMiller My query got resolved, actually I was originally trying to understand thermal diffusivity, then jumped upon heat capacity for understanding that. I read so many sources that got heavily confused, every source was stating something different, may be they meant the same, I wasn't getting the context. Started walking myself right from the basics of conduction and everything made sense. $\endgroup$ Commented Nov 4, 2021 at 20:02

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If we heat an amount $m$ of a substance with specific heat capacity $c_p$, from an initial temperaure $T_1$ to a final temperature $T_2$, then the amount of heat energy $Q$ needed (in the absence of any losses of course) is:

$$Q=\int_{T_1}^{T_2}m c_p(T)\mathrm{d}T$$

Assuming $m$ to be constant and $c_p(T)$ invariant to $T$ the formula reduces to:

$$Q=mc_p(T_2-T_1)$$

We can say that an amount of heat energy $Q$ is stored in $m$ mass of that medium.

We can also say that the storage capacity $c_p$ is given by:

$$c_p=\frac{Q}{m(T_2-T_1)}$$

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    $\begingroup$ Gert, a minor point. When you use $c_p$ you are assuming the object is an incompressible solid, which is what I did. For a gas, we need to know if the process is a constant volume or constant pressure process, i.e., whether $c_p$ or $c_v$ applies $\endgroup$
    – Bob D
    Commented Nov 2, 2021 at 23:08
  • $\begingroup$ @BobD Point taken, Bob. $\endgroup$
    – Gert
    Commented Nov 3, 2021 at 1:40
  • $\begingroup$ @Gert I have made an edit to the question, please help if you can. $\endgroup$ Commented Nov 4, 2021 at 13:31
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    $\begingroup$ @HarshitRajput Your edit is in fact a whole different question. You need to study Fourier's Heat Equation. This isn't something that can be edited in my answer. Sorry. en.wikipedia.org/wiki/Heat_equation $\endgroup$
    – Gert
    Commented Nov 5, 2021 at 10:40
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Consider 1g of iron, which has a specific heat of 0.45 J/g$^\circ $C. After heating it by 10 degrees, it stores an additional 4.5J of energy. You can compare this to other forms of potential energy, like compressing a spring, or raising a weight to some height, or charging a battery - you put in energy which gets stored in some way.

Contrast that with 1g of water, which has a specific heat of 4.2 J/g$^\circ $C - after heating by 10 degrees, it stores an additional 42J of energy. More energy is added than by raising the same mass of iron by the same number of degrees. This is analogous to compressing a stiffer spring, or raising a heavier weight, or charging a bigger battery - you put in more energy which gets stored in some way. For any particular mass and temperature, water holds more thermal energy than iron.

As a consequence of this, iron will heat up and cool off faster than water, since it takes less energy to go through the same temperature change. You could think of a cooling mass as a discharging battery, where stored heat energy is put back into the environment. A block of iron will cool relatively quickly, analogous to a battery that doesn't hold much charge, while the same mass of water will take much longer to cool, like a bigger battery that holds more charge.

In your added example of blocks A and B, we define the blocks to have the same mass but different heat capacities. Heat capacity is the amount of energy required to change the temperature of a unit mass of material. So, if you impart the same amount of energy to objects of different heat capacities, they will change in temperature by a different amount. As shown above, if you apply 42J of energy to 1g of water, it will increase in temperature by 10 degrees. If you apply that same amount of energy to 1g of iron, it will increase in temperature by 93 degrees. We've defined the situation where we impart the same amount of energy to both, in this setup, we're not "leaving behind" any heat by definition.

But suppose we don't impart the same amount of energy to both, but we instead just leave both masses in an oven at a specific temperature. Now we're not imparting the same energy to both, but we are instead going through the same temperature change in both. Since a higher heat capacity requires more energy to change temperature, the water absorbs more heat energy than the iron when going through the same temperature change. The iron will heat up quickly and reach the temperature of the oven, at which point it will not absorb any more heat energy or change in temperature. The water will continue to absorb heat, slowly increasing in temperature until it reaches the same temperature as the oven. At this point, both the iron and water are at the same temperature, but have absorbed different quantities of heat.

For objects of different heat capacity, you can either impart the same amount of energy for different temperature changes (e.g. by leaving both on identical bunsen burners for 1 second), or go through the same temperature change by imparting different amounts of energy (e.g. by leaving both in the oven until equilibrium is reached).

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  • $\begingroup$ I have made an edit to the question, please help if you can. $\endgroup$ Commented Nov 4, 2021 at 13:31
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It's because the substance must also release a quantity of energy to cool down and that quantity depends on the specific heat capacity.

For example a cylinder holding 100 kg of water at $80 C$ must release

$$100 \times 4200 \times 60 = 25.2 MJ$$

to cool to $20C$

Since the specific heat capacity of water is high, it makes water good at storing energy, in the sense that a smaller mass is needed to store energy for a given temperature difference, compared to other substances.

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  • $\begingroup$ I have made an edit to the question, please help if you can. $\endgroup$ Commented Nov 4, 2021 at 13:32
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It is the objects heat capacity $C$, which is the product of its specific heat $c$ and mass $m$ or $C=mc$, that relates to "heat storage" so that heat $Q=C\Delta T$. I've put heat storage in quotes because heat is not something that is stored. "Heat storage" is a common misnomer. Heat $Q$ is energy transfer due to solely to temperature difference. In the absence of work $W$, the energy transferred to the object by heat increases its internal energy, $\Delta U$, in accordance with the first law.

$$\Delta U=Q-W$$

Hope this helps.

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  • $\begingroup$ I have made an edit to the question, please help if you can. $\endgroup$ Commented Nov 4, 2021 at 13:31
  • $\begingroup$ @HarshitRajput Since you say blocks A and B start at the same temperature, there can be no heat transfer between them. The fact that the specific heat of one is greater than the other is irrelevant. Heat transfer requires a temperature difference. $\endgroup$
    – Bob D
    Commented Nov 4, 2021 at 13:50
  • $\begingroup$ sorry, I didn't mention that the blocks were not in contact, they're separate. Made the necessary changes. $\endgroup$ Commented Nov 4, 2021 at 15:25
  • $\begingroup$ If you transfer the same amount of heat to both blocks how can one store more energy than the other? A’s temperature will rise less than B since its specific heat is higher, $\endgroup$
    – Bob D
    Commented Nov 4, 2021 at 15:48
  • $\begingroup$ In the sense that Heat transfer is taking place from the bodies also. If only heat transfer were to happen to the bodies with no energy going out from them then they would've stored same amount of energy. $\endgroup$ Commented Nov 4, 2021 at 15:57
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In freshman physics, they taught us the first definition that you presented. However, that definition was applied exclusively to cases where the volume of the sample was constant and no work was done on- or by the sample. If work is done, the heat added is not equal to the specific heat at constant volume times the mass times the temperature change.

To make matters worse, the specific heat is supposed to represent an intrinsic physical property of the material (a function of state), while the amount of heat varies with the process path if work is done. To overcome these difficulties, in thermodynamics, we adopt a new approach to defining specific heat. Instead of defining it in terms of the amount of heat added, we define it in terms of the material's ability to store energy. This is done by basing it on the specific internal energy of the material (total microscopic molecular kinetic energy plus potential energy of molecular interactions per unit mass) U. So, according to this, the specific heat capacity at constant volume of the material is given by $$C_v=\left(\frac{\partial U}{\partial T}\right)_V$$ This is clearly a physical property of the material and, importantly, in addition, it reduces to the old freshman definition of specific heat when the volume is constant. However, this new thermodynamics definition is much more general and powerful than the old definition.

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  • $\begingroup$ I have made an edit to the question, please help if you can. $\endgroup$ Commented Nov 4, 2021 at 13:31
  • $\begingroup$ I don't understand the situation you are describing. Are you saying that A and B are in contact, and you add 10 kJ of heat to the side of A, and then let the two blocks equilibrate? $\endgroup$ Commented Nov 4, 2021 at 14:40
  • $\begingroup$ Oh sorry I didn't mention that. I've made the necessary changes. $\endgroup$ Commented Nov 4, 2021 at 15:24

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