Latent Heat

It is the energy required to change a substance's state without changing its temperature, like when ice melts or water boils. Discovered by Joseph Black in the 18th century, it comes in two types: latent heat of fusion (solid to liquid) and latent heat of vaporization (liquid to gas). The energy goes into breaking particle bonds, not increasing temperature. This concept is key in understanding weather, engines, and refrigeration.

1.0Introduction Latent Heat

• The word "latent" comes from the Latin term "latere," which means "to be hidden" or "concealed."
• Heat transfer usually changes a substance's temperature, but during a phase change, heat flow doesn't affect the temperature.
• All phase changes involve a change in internal energy, and the energy needed for this process is referred to as the heat of transformation.

2.0Definition of Latent Heat

• Heat required to change the state of unit mass depends on the nature of the substance and is called latent heat of substance.

$L=\frac{Q}{m}\Rightarrow Q=mL$

• When phase of a body changes, change of phase takes place at constant temperature [melting point or boiling point] and heat released or absorbed is Q = mL where L is latent heat.
• Heat is absorbed when a solid changes into a liquid (at the melting point) or when a liquid transforms into vapor (at the boiling point). Conversely, heat is released when a liquid freezes into a solid or when vapor condenses into a liquid.
• Phase Conversion : During the phase change, there will be no change in temperature. Whatever heat you provide, will go in changing the phase of the substance.

$Unit-S.I.\to J/Kg;CGS\to cal/g$

3.0Types of Latent Heat

1. Latent heat of fusion

• It is the amount of heat (in kilocalories) needed to transform 1 kg of a substance from its solid state to its liquid state at the melting point.
• Latent heat of fusion for ice $(L_f{)}_{\text{Ice}}=80\text{ cal/g}=80\text{ kcal/kg}$
• Latent heat of fusion for water $(L_f{)}_{\text{water}}=80\text{ cal/g}=80\text{ kcal/kg}$

$[Q=m{L}_f]$

2. Latent heat of vaporization

• The magnitude of heat needed to convert 1 kg of a substance from its liquid state to its vapor state at the boiling point.
• Latent heat of vaporization for water: $536\text{ kcal/kg}=536\text{ cal/g}\approx 540\text{ cal/g}$

$Q=m{L}_f\Rightarrow \left(L_v=\frac{Q}{m}\right)$

Note:

Example: Find the amount of heat required to convert 5 grams of ice from −10° C to steam at 130°C.

Solution:

$Q=ms\Delta T_1+m{L}_f+ms\Delta T_2+m{L}_v+ms\Delta T_2$

$Q=m\left(s_s\Delta T_1+L_f+s_w\Delta T_2+L_v+s_s\frac{\Delta T_3}{2}\right)$

$Q=5\left(\frac{1}{2}(10)+80+1(100)+540+\frac{1}{2}(30)\right)=3700\text{ cal}$

4.0Heating Curve

Heating of Solid:

When heat is supplied at a constant rate to a given mass (m) of a solid, and a graph is plotted between temperature and heat, the resulting graph is known as a heating curve.

Assumptions for heating curve

1. Initially the object is in solid state and its temperature is less than melting point.

2. Heat is given to the system by a constant power source.

$P=\frac{Q}{t}=\text{constant}$

$\text{Slope}=\frac{dy}{dx}=\frac{dT}{dQ}$

$dQ=msdT,\frac{dT}{dQ}\times \frac{1}{ms}=\text{slope}\Rightarrow ms=\frac{1}{\text{slope}}$

In the heating curve the inverse of slope gives heat capacity (ms).

• In the region OA, BC, DE, As the temperature of the liquid increases, the specific heat (or thermal capacity) of the liquid will be inversely proportional to the slope of the line.

$(\text{Heating Capacity}{)}_{\text{vapor}}>(\text{Heating Capacity}{)}_{\text{liquid}}>(\text{Heating Capacity}{)}_{\text{solid}}$

$\text{Latent heat}\propto \text{Length of graph parallel to x -axis}$

$L_f\propto \text{Length of line AB};L_v\propto \text{Length of line CD}$

• The latent heat of fusion is directly proportional to the length of the line (AB) with zero slope.

$\left[\text{In this region}-\text{specific heat}\propto \frac{1}{\tan 0^{\circ }}=\infty \right]$

• Latent heat of vaporisation is proportional to the length of line (CD) of zero slope.

$\left[\text{In this region}-\text{specific heat}\propto \frac{1}{\tan 0^{\circ }}=\infty \right]$

• If heat is supplied at a constant rate to a given mass (m) of a solid, and a graph is plotted between temperature and time, it would show the relationship between the two variables.

$Q=ms\Delta T\Rightarrow P\Delta t=ms\Delta T[:Q=P\Delta t]$

$\frac{\Delta T}{\Delta t}=\text{the slope of temperature-time curve}=\frac{1}{ms}$

5.0Solved Examples

1. The graph below illustrates the temperature change of a 5 kg substance as it absorbs heat at a constant rate of 42 kJ per minute. Find The latent heat of vaporization of the substance?       

Solution:

Liquid to gas conversion takes place in between [20 min to 30 min]

Total time = 10 min

$\Delta Q=(42\times 10^3)(10)=42\times 10^4$

$\Delta Q=m{L}_f$

$L_f=\frac{42\times 10^4}{5\text{ kg}}=84\text{ kJ/kg}$

2. A calorimeter with a water equivalent of 20 g contains 5 g of ice at 0°C. How much heat is required to completely evaporate the water that results from melting the ice? (Assume no heat loss.)

Solution: 

$Q=m{L}_f+m{s}_w\Delta T_2+m{L}_v+Ms\Delta T\Rightarrow Q=m\left(L_f+s_w\Delta T_2+L_v\right)+W\Delta T$

$Q=5(80+1(100)+540)+20(100-0)\Rightarrow Q=3600+2000=5600\text{ cal}$