# Transferred heat: formulas, how to calculate it and solved exercises

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

A good example is a steaming cup of coffee. The metal teaspoon with which the sugar is stirred is heated. If left in the cup long enough, the coffee and the metal teaspoon will eventually match their temperatures: the coffee has cooled and given heat to the teaspoon. Some heat has passed into the environment as the system is not insulated.

When temperatures are equalized, *thermal equilibrium has* been reached.

If the same test were done with a plastic spoon, you would certainly find that it doesn’t heat up as quickly as metal, but eventually it will also balance out with the coffee and everything around it.

This is because metal conducts heat better than plastic. On the other hand, coffee certainly produces heat at a different rate than hot chocolate or another beverage. So, the heat supplied or absorbed by each object depends on what material or substance is made.

__what is it and formulas__

__what is it and formulas__

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

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

Usually Q is called the amount of heat the hottest object produces. This value is proportional to the mass of that object. A body with a large mass is capable of providing more heat than one with a smaller mass.

**The temperature difference ***ΔT*

*ΔT*

Another important factor in calculating the assigned heat is the temperature difference experienced by the object providing the heat. It is denoted as Δ *T* and is calculated as:

*ΔT = T _{f} – T _{or}*

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

So, finally, the expression for transferred heat is as follows:

*Q _{produced} = – mcΔ *

*T*

Income is symbolized with a minus sign.

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

Specific heat is the amount of heat needed to raise the temperature of 1 g of 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).

Thermal capacity C is a linked concept, but a little different in that the object’s mass intervenes. The heating capacity is defined as follows:

*C = mc*

Its SI units are Joule / K. Therefore, transferred heat can also be expressed equivalently to:

*Q = -C. Δ **T*

__How to calculate?__

__How to calculate?__

To calculate the heat provided by an object, you need to know the following:

– The specific heat of the substance that produces heat.

– mass of said substance

– The final temperature to obtain

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

**Calorimetry**

However, if this value is not known, it is possible to obtain it with the help of a thermometer and water in a container with thermal insulation: the calorimeter. A schematic of this device is shown in the figure accompanying exercise 1.

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

By comparing the result with the tabulated values, you can know what substance it is. This procedure is called *calorimetry.*

The heat balance is carried out saving energy:

Q _{assigned} + Q _{absorbed = 0}

**solved exercises**

**Exercise 1**

A 0.35 kg piece of copper is introduced at a temperature of 150 °C into 500 mL of water at a temperature of 25 °C.

a) The final equilibrium temperature

b) How much heat flows in this process?

**Data**

*c _{copper} = 385 J / kg. °C*

*c _{water =} 4180 J / kg. °C*

*Water density: 1000 kg / m ^{3}*

**Solution**

a) Copper produces heat while water absorbs it. As the system is considered closed, only the water and the sample intervene in the heat balance:

*Q _{assigned} = Q _{absorbed}*

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

*500 mL = 0.5 L = 0.0005 m ^{3}*

With these data, the water mass is calculated:

*= Mass x volume density = 1000 kg / m ^{3} . 0.0005 m ^{3} = 0.5 kg*

The equation of heat in each substance is presented:

*Q _{produced} = -m _{copper} . c _{copper} . AT *

*= -0.35 kg. 385 J/kg.*

*(T*

_{f}–150 °C) = -134.75 (T_{f}– 150) J*Q _{absorbed} = m of _{water} . c _{water} . AT *

*= 0.5 kg. 4186 J/kg. °C. (T*

_{f}–25 °C) = 2093 (T_{f}–25) JCorresponding to the results you have:

*2093 (T _{f} – 25) = -134.75 (T _{f} – 150)*

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

*T _{f} = 32.56 °C*

b) The amount of heat that flows is the heat transferred or the heat absorbed:

Q _{produced} = – 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 an oven at a temperature of T _{o} and then introduced into a 150 g copper calorimeter containing 200 g of water at 16 º C. The final temperature, once balanced, is 38 º C. When the calorimeter and its contents are weighed, 1.2 g of water is found to have evaporated. What was the initial temperature T _{o} ?

*Data: the latent heat of water vaporization is L _{v} = 2257 kJ / kg*

**Solution**

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

– Heat the water calorimeter (200 g)

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

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

*Q _{produced}* = –

*100*

*x*

*1 x 10*= –

^{-3}kg. 385 J / kg °C. (38 – T_{o}) ºC*38.5*.

*(38 – T*

_{or}) J*Q _{absorbed by }_{calorimeter} = Q _{absorbed by water} + Q _{vaporization} + Q _{absorbed by copper}*

*0.2 kg 4186 J / kg °C. (38 – 16 ºC) + 1.2 x 10 ^{-3} kg. *2257000 J / kg

*+0.150 kg .385 J / kg. ° C (38 – 16 ° C) =*

*18418.4 +2708.4 + 1270.5 J = 22397.3 J*

Therefore:

– *38.5* . *(38 – T _{o} ) = 22397.3*

*T _{o} = 619.7 °C*

The heat required to bring 1.2 g of water to 100°C may also have been considered, but it’s a pretty small amount in comparison.