**Heat**

When two objects have different temperature, energy is transferred from one object to the other, this energy is called as **heat**.

Heat always flow from objects with higher temperature to the object with lower temperature. An object does not contain heat, but internal energy. Internal energy refers to all the energy contained in a substance when viewed from a reference frame at rest with respect to the center of mass of the system. That is, when an object is being heated, the molecules move faster, increasing the kinetic energy of the object, thus, making its temperature higher. In this case, the object is said to have gained internal energy. Therefore, heat is then considered to be a measure of the change in the total internal energy of a body. Internal energy is also dependent to the amount of the substance. The greater amount of the substance, the higher the internal energy.

To measure the amount of heat being transferred, the changes that happened on a system during the process like the increase or decrease in temperature, is being taken in consideration. The most commonly used unit for heat is the **calorie** – which is the amount of heat required to change the temperature of 1 gram of water by 1 ºC. In the U.S. customary system, the unit used is the **British thermal unit (Btu)** which is defined as the amount of heat needed to raise the temperature of 1 lb of water from 63 °F to 64 °F. Another unit used is the **joule**, which is used when mechanical processes are involved.

**Mechanical Equivalent of Heat**

James Prescott Joule, an English physicist, discovered a relationship between heat and mechanical energy. He found out that these 2 quantities are inconvertible, and the transformations from one to the other is in fixed proportion, which is known as the mechanical equivalent of heat. In his experiment, Joule found that the decrease in mechanical energy is proportional to the product of the mass of the water and the increase in the water’s temperature. This proportionality constant was approximately equal to 4.18 J/g ºC. This means that, 4.18 J of mechanical energy is needed to raise the temperature of a 1 gram of water by 1 ºC. Thus, the mechanical equivalent of heat is:

**1 cal = 4.186 J**

**Specific Heat**

The amount of heat needed to change or raise the temperature of certain substances may differ. A substance may take lesser amount of heat to raise its temperature compared to the same substance but with larger mass. The amount of heat needed to raise the temperature of a substance by 1 ºC or 1 K is called as the **heat capacity (c)**. It can be represented by the equation

\(\mathbf{c = {Q \over \Delta T}}\)

where ** c** is the heat capacity,

**is the amount of heat absorbed or the energy that changes the temperature of the substance, and**

*Q**\(\mathbf{\Delta T}\)*is the change in temperature.

Since the heat capacity depends on the substance’s mass, the amount of heat needed to change the temperature per mass of the object can also be solved. This is called as the ** specific heat capacity** or

**– which is defined as the amount of heat required to change the temperature of each unit mass of a particular substance by 1 degree. Mathematically, specific heat can be expressed as**

*specific heat*\(\mathbf{c = {\Delta Q \over m\Delta T}}\)

where ** c **is the specific heat capacity of the substance in cal/g ºC or J/kg K, \(\Delta Q\) is the amount of heat absorbed or released,

**is the mass of the substance, and \(\Delta T\) is the change in temperature.**

*m***Example 1.**

How much heat is needed to raise the temperature of a 0.90 kg ethyl alcohol from 20 ºC to 70 ºC? The specific heat of ethyl alcohol is 0.60 cal/g·ºC.

Given:

m = 0.90 kg = 900 g

T_{i} = 20 ºC

T_{f} = 70 ºC

c = 0.60 cal/g·ºC

Solution: First, we have to derive the formula for the amount of heat needed from the speicific heat equation:

\(c = {\Delta Q \over m\Delta T} \implies \Delta Q = mc(T_f – T_i)\)

Using the equation above and substituting the given values, we have

\(\Delta Q = (900\;g)(0.60\;cal/g\cdot ^\circ C)(70 ^\circ C – 20^\circ C) = 27, 000\;cal\).

Thus, the amount of heat needed to raise the temperature of a 0.90 kg ethyl alcohol from 20 ºC to 70 ºC is ** 27, 000 calories**.

**Example 2.**

The specific heat of aluminum is 0.217 cal/g·ºC. What amount of aluminum gives off -434 cal of heat if it is cooled down from 100.0 ºC to 20.0 ºC?

Given:

Q = -434 cal

T_{i} = 100.0 C

T_{f} = 20.0 C

c = 0.217 cal/g·ºC

Solution:

Derive the formula for mass from the specific heat equation:

\(c = {\Delta Q \over m\Delta T} \implies m = {\Delta Q \over c \Delta (T_f – T_i)}\)

Substitute all the given values to the equation:

\(m= {-434\; cal \over {(0.217\;cal/g \cdot ^\circ C)(20.0\;^\circ C – 100.0\;^\circ C)}}=25.0\;g\)

Therefore, * 25.0 grams *is the amount of aluminum that gives off -434 cal of heat if cooled down from 100.0 ºC to 20.0 ºC