Specific latent heat and specific heat capacity

In a nutshell

The specific heat capacity of a substance is the energy per unit mass needed to change its temperature by $1\ K$ . The specific latent heat of a substance is the energy needed per unit mass to change its phase.

Equations

description	equation
Specific heat capacity equation	$E=mc\Delta\theta$
Specific latent heat equation	$E=mL$

Variable definitions

quantity name	symbol	derived units	alternative units	si base units
$energy$	$E$	$J$	$eV$	$kg\ m^2\ s^{-2}$
$mass$	$m$	$kg$		$kg$
$specific\ heat\ capacity$	$c$	$J\ kg^{-1}\ K^{-1}$		$m^2\ s^{-2}\ K^{-1}$
$temperature$	$T/\theta$	$K$	$\degree C$	$K$
$specific\ latent\ heat$	$L$	$J \ kg^{-1}$		$m^2\ s^{-2}$

Specific heat capacity

The specific heat capacity of a substance is defined as the energy needed to change the temperature of $1\ kg$ of mass of a substance by $1\ K$ . It is symbolised by the letter $c$ and has units $J\ kg^{-1}\ K^{-1}$ .

The energy needed to change the temperature of a substance can be found with the equation:

$E=mc\Delta\theta$

Note: The $\Delta$ means change in the variable in front of it. Also, when you are finding the change in temperature, you can keep the temperature in $\degree C$ as the change in temperature will be the same in $K$ .

The specific heat capacity of a substance changes depending on its phase. For example, the specific latent of water is different to that of ice.

Specific heat capacity at thermal equilibrium

When there are two materials that are in equilibrium, you can equate the specific heat capacity equation of each substance. This is useful when working out an unknown mass or specific heat capacity.

Example

A steel spatula is used to cook dinner and reaches a temperature of $100 \ \degree C$ . The spatula is then placed in $2.0 \ kg$ of water at $35 \ \degree C$ . Eventually the water and the spatula reach thermal equilibrium at $40 \ \degree C$ .

Calculate the mass of the spatula. Assume all of the spatula is steel and submerged in the water. The specific heat capacity of steel and water respectively is $420 \ J \ kg^{-1} \ K^{-1}$ and $4200\ J\ kg^{-1}\ K^{-1}$ .

Firstly, write down everything you know:

$c_{steel} = 420 \ J \ kg^{-1} \ K^{-1} \\ \Delta \theta_{ steel} = 100 \degree C - 40 \degree C = 60 \degree C \\ c_{water} = 4200 \ J \ kg^{-1} \ K^{-1} \\ m_{water} = 2.0 \ kg \\ \Delta \theta_{water} = 40 \degree C - 35 \degree C = 5 \degree C$

Next, write down the equation needed:

$E=mc\Delta\theta$

Equate the specific heat capacity equation for water and steel:

$m_{steel}c_{steel}\Delta\theta_{steel} = m_{water}c_{water}\Delta\theta_{water}$

Rearrange for $m_{steel}$ by dividing by $c_{steel}\Delta\theta_{steel}$ :

$m_{steel} = \dfrac{m_{water}c_{water}\Delta\theta_{water}}{c_{steel}\Delta\theta_{steel}}$ 

Substitute in the values and calculate the final answer:

$m_{steel} = \dfrac{2.0 \times 4200 \times 5}{420 \times 60} = 1.66667 ... \ kg$

The mass of the steel spatula is $\underline{1.7 \ kg}$ to 2 significant figures.

Specific latent heat

The specific latent heat is defined as the energy needed to change the phase of $1\ kg$ of a substance while at constant temperature. It is symbolised by the letter $L$ and has units $J\ kg^{-1}$ .

The energy needed to change the phase of a substance can be found with the equation:

$E=mL$

There are two types of specific heat capacity which depend on the phase change:

When the substance changes from solid to liquid it is referred to as specific latent heat of fusion $L_f$ .
When the substance changes from liquid to gas it is referred to as specific latent heat of vaporisation $L_v$ .

Example

Calculate the total energy required to turn a $2\ kg$ block of ice at $-5\degree C$ into $20\degree C$ water. The specific heat capacity of ice and water are $2000\ J\ kg^{-1}\ K^{-1}$ and $4200\ J\ kg^{-1}\ K^{-1}$ respectively, while the specific latent heat of fusion for ice is $3.3\times10^{5}\ J\ kg^{-1}$ .

Firstly, write down everything you know:

$m=2\ kg \newline \theta_{initial}=-5\degree C\newline \theta_{final}=20\degree C\newline c_{ice}=2000\ J\ kg^{-1}\ K^{-1}\newline c_{water}=4200\ J\ kg^{-1}\ K^{-1}\newline L_{f\ ice}=3.3\times10^{5}\ J\ kg^{-1}$

Next, write down all the steps and phase changes it takes for the ice to become water:

$ice(-5\degree C)\xrightarrow{heat} ice(0\degree C)\xrightarrow{phase\ change} water(0\degree C)\xrightarrow{heat} water(20\degree C)$ 

The total energy required to turn the ice into water is the sum of the energies needed to heat and to change phase:

$E_{tot}=E_{ice(-5\degree C\rightarrow 0\degree C)}+E_{fusion}+E_{water(0\degree C\rightarrow20\degree C)}$ 

Write down the equations for the energy needed to change temperature and to change phase:

$E=mc\Delta\theta\newline E=mL$

Substitute them into the equation for the total energy:

$E_{tot}=mc_{ice}\Delta\theta_{(-5\degree C\rightarrow0\degree C)}+mL_{f\ ice}+mc_{water}\Delta\theta_{(0\degree C\rightarrow20\degree C)}$ 

Calculate the differences between the temperatures:

$\Delta\theta_{(-5\degree C\rightarrow0\degree C)}=5\degree C \newline \Delta\theta_{(0\degree C\rightarrow20\degree C)}=20\degree C$ 

Substitute all the numbers into the equation:

$E_{tot}=2\times2000\times5+2\times3.3\times10^{5}+2\times4200\times20$ 

Calculate the total energy:

$E_{tot}=8.48\times10^{5}\ J=848\ kJ$ 

The total energy needed to turn a $2\ kg$ block of $-5\degree C$ ice into water at $20\degree C$ is $\underline{800\ kJ}$ to one significant figure.

Note: Remember to round your answers to the lowest number of significant figures given.