Heat is a transfer of energy because of temperature difference. When an object gains or loses heat it either changes its temperature or its phase (state). A change in temperature is associated with changes in the average kinetic energy of the particles and a change in phase is associated with changes in the internal potential energy possessed by the object. In this section we will learn how to measure the quantity of heat gained or released by an object.
Subsubsection4.1.1.1Specific Heat
Heat absorbing capacity of a material is known as specific heat capacity (or simply, specific heat). It is a material property. Different materials absorb different amounts of heat even for the same rise of temperature. For example, if you heat an identical iron ball and an aluminum ball in the same pot for the same time duration, then you would be surprised that the aluminum ball is colder than the iron ball. This is because the specific heat of iron is less than that of aluminum. In iron, more of the heat energy is spent to increase the kinetic energy of its molecules than to increase the intermolecular potential energy. Hence, the iron has a higher temperature. The quantity of total heat transferred to the body is given by the following formula:
Where, \(Q\) stands for the amount of heat required to change the temperature, \(m\) stands for the mass of the substance, \(c\) stands for the specific heat capacity of the substance, and \(\Delta T\) stands for the temperature change (i.e., \(\Delta T=T_f-T_i\)). Hence specific heat can be defined as the amount of heat required to change the temperature of 1 kg of the substance to \(1 ^oC\text{.}\) Its SI unit is \(J/kg\cdot^oC\text{.}\) The greater the specific heat of the substance, the greater would be the amount of heat needed to raise the temperature per unit mass. The higher the specific heat of the substance, the higher the amount of heat it can hold. The table below shows the specific heat of some common materials.
Table4.1.3.Specific heats of some common substances
Substance
Specific heat (\(cal/g/^oC\))
Water
1
Sand
0.16
iron
0.1
aluminum
0.22
Wood
0.42
Ice
0.5
Example4.1.4.
How much heat in calorie is needed to heat 50 kg of water from \(15 ^oC\) to \(40 ^oC\text{.}\)
Solution.
Given: m = 50 kg =50000 g, \(T_i=15 ^oC\text{,}\)\(T_f=40 ^oC\text{,}\)\(c=1cal/g.^oC\text{,}\) and Q=?
Heat absorbed or released by the material during phase change is called latent heat. There is no temperature change during a phase change hence this energy is hidden from temperature scale. The energy dissipated or released during the phase change is in adjustment of particles potential energy in the bonds. Hence the amount of heat \(Q\) during phase change is given by
\begin{equation}
Q = mL \tag{4.1.3}
\end{equation}
Here \(Q\) is the amount of heat required to change the state or phase of the substance at change of state temperature, \(m\) is the mass of the substance, and \(L\) is the latent heat of the substance. Latent heat is also a material property. The latent (hidden) heat of fusion for water at \(0 ^oC\) is approximately 79.7 calories (334 J) per gram, and the latent heat of vaporization at \(100 ^oC\) is about 533 calories (2,230 joules) per gram. Simulation for Boiling Water 2
