He **given up heat** It is the transfer of energy between two bodies at different temperatures. The one with the highest temperature gives up heat to the one whose temperature is lower. Whether a body gives up or absorbs heat, its temperature or its physical state can vary depending on the mass and the characteristics of the material from which it is made.

A good example is a steaming cup of coffee. The metal spoon with which the sugar is stirred is heated. If left in the cup long enough, coffee and metal spoon will eventually equalize their temperatures: the coffee will have cooled and it will have given up heat to the spoon. Some heat will have passed to the environment, since the system is not isolated.

When the temperatures were equalized, the *thermal equilibrium*.

If you did the same test with a plastic teaspoon, you’d probably notice that it doesn’t heat up as quickly as the metal one, but eventually it will also come into equilibrium with the coffee and everything around it.

This is because metal conducts heat better than plastic. On the other hand, surely coffee gives off heat at a different rate than hot chocolate or another drink. So the heat given up or absorbed by each object depends on what material or substance it is made of.

[toc]

**What it consists of and formulas**

Heat always refers to the flow or transit of energy between one object and another, due to the difference in temperature.

That is why we speak of transferred heat or absorbed heat, since by adding or extracting heat or energy in some way, it is possible to modify the temperature of an element.

Normally, Q is the amount of heat that the hottest object gives off. This value is proportional to the mass of said object. A body with great mass is capable of giving up more heat than one of less mass.

**temperature difference ***ΔT*

*ΔT*

Another important factor in calculating the heat released is the temperature difference experienced by the object giving off the heat. It is denoted as Δ*you* and it is calculated like this:

*ΔT= Tf – To*

Finally, the amount of heat released also depends on the nature and characteristics of the object, which are quantitatively summarized in a constant called *specific heat of material*denoted as *c*.

So finally the expression for the heat transferred is the following:

*Qceded = – mcΔ**you*

The fact of yielding is symbolized by a negative sign.

**The specific heat and heat capacity of a substance**

Specific heat is the amount of heat required to raise the temperature of 1 g of a substance by 1 °C. It is an intrinsic property of the material. Its units in the International System are: Joule/kg. K (Joule between kilograms x temperature in degrees Kelvin).

The heat capacity C is a related concept, but slightly different, since the mass of the object is involved. Heat capacity is defined as follows:

*C= mc*

Its SI units are Joule/K. So the heat lost can also be expressed equivalently as:

*Q = -C. Δ**you*

**How to calculate it?**

To calculate the heat released by an object, it is necessary to know the following:

– The specific heat of the substance that gives up the heat.

– The mass of said substance

– The final temperature to be obtained

Specific heat values for many materials have been determined experimentally and are available in tables.

**Calorimetry**

Now, if this value is not known, it is possible to obtain it with the help of a thermometer and water in a thermally insulated container: the calorimeter. A schematic of this device is shown in the figure that accompanies exercise 1.

A sample of the substance at a certain temperature is immersed in a quantity of water that has previously been measured. The final temperature is measured and with the values obtained the specific heat of the material is determined.

Comparing the result with the tabulated values, it can be known what substance it is. This procedure is called *calorimetry.*

The heat balance is carried out by conservation of energy:

Q ceded + Q absorbed = 0

**solved exercises**

**Exercise 1**

A 0.35 kg piece of copper at a temperature of 150 ºC is introduced into 500 mL of water at a temperature of 25 ºC. Find:

a) The final equilibrium temperature

b) How much heat flows in this process?

**Data**

*ccopper =385 J/kg . ºC*

*water = 4180 J/kg . ºC*

*Water density: 1000 kg/m3*

**Solution**

a) Copper gives off heat while water absorbs it. Since the system is considered closed, only the water and the sample are involved in the heat balance:

*Q ceded = Q absorbed*

On the other hand, it is required to calculate the mass of 500 mL of water:

*500 mL = 0.5 L = 0.0005 m3*

With these data the mass of water is calculated:

*mass = density x volume = 1000 kg/m3 . 0.0005 m3 = 0.5 kg*

The equation for the heat in each of its substances is posed:

*Qceded = -mcobre . copper. Δ**T =-0.35kg. 385 J/kg .ºC . **(Tf –150 ºC) =-134.75 (Tf – 150) J*

*Qabsorbed =magua . cagua. Δ**T = 0.5kg. 4186 J/kg . ºC .(Tf –25 ºC) = 2093 (Tf –25) J *

Equating the results, we get:

*2093 (Tf – 25) = -134.75 (Tf – 150)*

It is a linear equation with one unknown, whose solution is:

* **Tf = 32.56 ºC*

b) The quantity of heat that flows is either the heat given up or the heat absorbed:

Q ceded = – 134.75 (32.56 – 150) J = 15823 J

Q absorbed = 2093 (32.56 – 25) J = 15823 J

**Exercise 2**

A 100 g piece of copper is heated in a furnace to a temperature To and then introduced into a 150 g copper calorimeter containing 200 g of water at 16ºC. The final temperature once in equilibrium is 38º C. When the calorimeter and its contents are weighed, 1.2 g of water is found to have evaporated. What was the initial temperature To?

*Data: The latent heat of vaporization of water is Lv = 2257 kJ/kg*

**Solution**

This exercise differs from the previous one, since it is necessary to consider that the calorimeter also absorbs heat. The heat given up by the piece of copper is invested in all of the following:

– Heat the water in the calorimeter (200 g)

– Heat the copper from which the calorimeter is made (150 g)

– Evaporate 1.2 grams of water (energy is also needed for a phase change).

*ceded* = –*100 **x** 1 x 10 -3 kg. 385 J/kg . ºC. (38 – To )ºC* = –*38.5*. *(38 –To)J*

*Q absorbed by the calorimeter = Q absorbed by the water + Q vaporization + Q absorbed by the copper*

*0.2 kg .4186 J/kg ºC .(38 – 16 ºC) + 1.2 x 10-3 kg. *2257000 J/kg* +0.150 kg .385 J/kg .ºC.(38 – 16 ºC) =*

*18418.4 +2708.4 + 1270.5 J = 22397.3 J*

Therefore:

–*38.5*. *(38 – To)= 22397.3*

* **To = 619.7 ºC*

The heat needed to bring the 1.2 g of water up to 100ºC could also have been considered, but it is a rather small amount in comparison.

**References**

Giancoli, D. 2006. Physics: Principles with Applications. 6th. Ed. Prentice Hall. 400 – 410. Kirkpatrick, L. 2007. Physics: A look at the world. 6th Abridged Edition. Cengage Learning. 156 – 164. Rex, A. 2011. Fundamentals of Physics. pearson. 309-332. Sears, Zemansky. 2016. University Physics with Modern Physics. 14th. Ed. Volume 1. 556 – 553. Serway, R., Vulle, C. 2011. Fundamentals of Physics. 9th Cengage Learning.