If boiling of water involves change in internal energy, then why does the temperature remain constant? According to the first law of thermodynamics,
$$\Delta Q=\Delta W+\Delta U$$
Considering boiling of water to be an isothermal process, $\Delta U$ should be zero, but then my textbook says: "we see that most of the heat goes to increase the internal energy of water in transition from the liquid to the vapour phase"
So, I am actually really confused here, I have been taught that temperature represents the kinetic energies of the molecules of the system that is internal energy. In this case, the internal energy is clearly increasing, since we are transitioning from liquid to vapour phase, then why doesn't the temperature change?
Edit: This question really bugged me a few months ago and I thought I understood it but after all those months when I studied this topic again, I have stumbled on the same question. Reading previous comments and answers it still isn't clear to me. I will really appreciate it if someone would provide a definition as to what temperature really means because as far as I understand it, if internal energy is increasing, temperature should also increase (if we go with the definition that temperature is the manifestation of the average energy of molecules).
 A: There is also a variational approach to temperature. Your system is an open one, since it can exchange energy. Assuming for simplicity that volume is fixed, the second law is not about maximising entropy, $S$, anymore, but rather minimising the free energy, $F$:
$$
F = U-TS
$$
In this definition, $T$ measures the compromise between minimising energy and maximising entropy. At high temperature, the entropy wins over, at low temperature, energy wins over. This is how you can interpret phase transition, going from a low energy state to a high entropy one.
Even from a microscopic origin, this approach can define the canonical ensemble. From this perspective, temperature is a measure of the compromise the distribution makes between the uniform one (maximal entropy) and the one concentrated at the low energy micro states (minimal energy).
This can explain your problem as well. Equivalently, I have to explain why while maintaining temperature, $U$ can increase. If $U$ can increases, it increases $F$, which is immediately compensated by an increase in $S$. Actually, at phase coexistence, $F$ is constant, so you can say that energy is exactly "converted" to entropy. This approach is not only useful to forge intuition, but can be used to give quantitative results like the Clausius-Clapeyron relation.
Btw, in a more realistic setting, the system can exchange volume with the environment. This will rather lead to the consideration of free enthalpy $G = F+pV$ which should be more familiar if you've done chemistry or thermodynamics. The discussion is essentially the same though.
Hope this helps.
Answer to comment
The result is a general one for phase coexistence. Check out for example Shroeder’s Thermal Physics. The idea is that each phase has its own free energy. Once again, this kind of consideration is more usually done with free enthalpy (chemical potential), but in our case, this will do. I’ll write $F_l,F_g$ the respective free energies of the liquid phase of gas phase. If $F_l>F_g$ the system gains by being entirely in the gas phase. The converse is also true. So you have phase coexistence iff the free energies are equal $F_l=F_g$.
Note that I am assuming that $F_g$ (resp. $F_l$) is still defined below (resp. above) the boiling point. It actually makes sense. In fact, below (resp. above) the boiling point, the gas (resp. liquid) state is metastable, giving rise to supercooled vapor (resp. superheating).
At the boiling point, you still have a degree of freedom, the relative proportion of water and vapor which will determine $F$. When you add energy, to keep $F$ constant, the entropy increases. This means you convert some water to vapor, and this change of state absorbs the excess energy. Note that this works as long as you still have phase coexistence. If you add too much energy, you vaporize everything and temperature starts increasing again.
Mathematically:
$$
\Delta U=T\Delta S
$$
which is why I said that figuratively, energy is converted to entropy. In fact, these variations are proportional to the amount of water that changes state. This is why I don’t think that you can explain the phenomenon only using energy, at one point you’ll have to use entropy (though it can be disguised using microscopic arguments).
A: The average kinetic energy of a molecule is indeed proportional to temperature:
$${1\over 2}m\langle v^2\rangle={3\over 2}k_BT$$
In a liquid, the potential energy due to the interactions between molecules also contributes to the average energy. The interaction between molecules is attractive (usually modeled by a Lenard-Jones potential) so the potential energy is negative and the average energy is lower in the liquid phase.
When a liquid is heated, the thermal energy brought to the system first leads to an increase of the average velocity. At $100^\circ{\rm C}$, the free energy $F=E-TS$ of the gas phase becomes equal to the free energy of the liquid phase. Some molecules start to leave the liquid. However, in the gas phase, the average total energy of the molecules is higher than in the liquid phase. Molecules can therefore leave the liquid only if the energy difference is brought to the system under the form of thermal energy. This energy is used to escape from the attraction of other molecules. If the system is still heated, the fraction of molecules in the gas phase will smoothly increase (and therefore the fraction in the liquid phase will decrease). The kinetic energy remains the same but less and less molecules have a (negative) potential energy. As a consequence, the total energy of the system will increase, even though the temperature does not change. When all molecules will be in the gas phase then any further thermal energy brought to the system is converted into kinetic energy of the molecules and will lead again to an increase of the velocity of the molecules and of temperature.
During the liquid-gas phase transition, the free energy $F=E-TS$ is continuous but the energy $E$ is a discontinuous function of the temperature.
A: 
I will really appreciate it if someone would provide a definition as to what temperature really means because as far as I understand it, if internal energy is increasing, temperature should also increase (if we go with the definition that temperature is the manifestation of the average energy of molecules).

Temperature is always proportional to average kinetic energy of molecules, whether gas, liquid or solid.
Internal energy is kinetic energy plus potential energy. Potential energy has not simple relation to temperature.
When boiling water, adding heat does not increase temperature of the liquid, and thus does not increase average kinetic energy of a molecule. The added energy increases potential energy of the system liquid+vapor, as vapor molecules get farther from the liquid molecules. Internal energy increases, potential energy increases, kinetic energy stays the same.
A: Temperature characterizes the average energy of atoms/molecules, as per Boltzmann distribution:
$$
w(\mathbf{p},\mathbf{q})\propto e^{-\frac{H(\mathbf{p},\mathbf{q})}{k_B T}},$$
where the potential energy may contain interactions, i.e., it is not simply a sum of potential energies of separate atom/molecules in external field. Further, one can then prove the equipartition theorem that the average energy per degree of freedom is $k_B T/2$.
However, there is nothing in these statements about the temperature being the kinetic energy. This is the case when we consider an ideal gas, in which (by its definition) the interactions between the atoms/molecules are neglected and all the energy is the kinetic one.
Thus, the reason why the boiling water is not increasing its temperature, is because the energy goes into overcoming the potential keeping the molecules close to each other in the liquid phase.
Update (Correcting some of the misleading statements made above.)
Let us consider a liquid-gas phase transition for a system of $N$ particles described by Hamiltonian
$$
H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)=\sum_{i=1}^N\frac{\mathbf{p}_i^2}{2m} + \sum_{i=1}^N\sum_{j=1}^{i-1}U\left(|\mathbf{r}_i-\mathbf{r}_j|\right)=\sum_{i=1}^NK(\mathbf{p}_i) + V(\mathbf{r}_1;..., \mathbf{r}_N)
$$
The Boltzmann distribution is then
$$
w(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)=
Z^{-1}e^{-\frac{H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)}{k_BT}}= 
Z^{-1}
\left(\prod_{i=1}^Ne^{-\frac{\mathbf{p}_i^2}{2mk_BT}}\right)\times
e^{-\frac{V(\mathbf{r}_1;..., \mathbf{r}_N)}{k_BT}}
$$
The partition function is given by
$$
Z = \left(\prod_{i=1}^N\int d^3\mathbf{p}_ie^{-\frac{\mathbf{p}_i^2}{2mk_BT}}\right)\int_{\Omega^N}\prod_{i=1}^N d^3\mathbf{r}_i e^{-\frac{V(\mathbf{r}_1;..., \mathbf{r}_N)}{k_BT}}=
\left(4\pi mk_BT\right)^{3N}Z_1
$$
If we calculate the average kinetic energy of a single molecule, the configuration integral over space cancels out, and it reduces to simple Gaussian integration:
$$
\left \langle\frac{\mathbf{p}_i^2}{2m} \right\rangle=
\frac{\int d^3\mathbf{p}_i \frac{\mathbf{p}_i^2}{2m} e^{-\frac{\mathbf{p}_i^2}{2mk_BT}}}{\int d^3\mathbf{p}_ie^{-\frac{\mathbf{p}_i^2}{2mk_BT}}}=
3\left\langle\frac{p_{ix}^2}{2m}\right\rangle=
\frac{3\langle p_{ix}^2\rangle}{2m}=
\frac{3mk_BT}{2m}=\frac{3 k_BT}{2}
$$
In other words, the average kinetic energy of the molecules is proportional to the temperature.
The above remains true even at the temperature where the system undergoes phase transition, as the result above is independent in the configuration integral $Z_1$. It is this configuration integral that undergoes change in phase transition: in the gas phase the volume filled with gas is that of the containing reservoir, $\Omega_g$, whereas in the liquid phase it is only the volume filled with the liquid, $\Omega_l$ (note that this volume can be located in different parts of the container or even be composed of many disconnected volumes.) Thus, the internal energy is different in two phases, depending on the volume over which we integrate:
$$
\left\langle H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)\right\rangle=
\int\prod_{i=1}^N d^3\mathbf{r}_i\int_{\Omega^N}\prod_{i=1}^N d^3\mathbf{r}_i H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N) Z^{-1}e^{-\frac{H(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)}{k_BT}}=
\frac{3Nk_BT}{2} + Z_1^{-1}\int_{\Omega^N}\prod_{i=1}^N d^3\mathbf{r}_i V(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N) Z^{-1}e^{-\frac{V(\mathbf{p}_1, \mathbf{r}_1;...,\mathbf{p}_N, \mathbf{r}_N)}{k_BT}}.
$$
Thus, the internal energy experiences a discontinuity across the phase transition (see also Why is the internal energy discontinuous in a first-order phase transition?, note that we are dealing here with a first order phase transition.) This discontinuity equals to the latent heat absorbed or produced during the phase transition.
Of course, in practice energy does not change discontinuously, but gradually, as heat is transferred or taken away from the substance. However, in this process the substance is not in thermal equilibrium, and needs to be treated using the methods of non-equilibrium statistical physics, see, e.g., Phase Transition in the Boltzmann–Vlasov Equation.
Related question (brought up in the comments by @GiorgioP): Why, exactly, does temperature remain constant during a change in state of matter?
A: Here is a thermodynamics (macro) answer that may help.  Others have provided a statistical mechanics (micro) answer. I added a short discussion of Evaporation in general, and two specific types of evaporation: Boiling and Flashing.
Your first law relationship is only true for a closed system, one with no mass transfer in/out of the system.  For an open system $\Delta U = \Delta Q - \Delta W + \Delta H$ where $\Delta H$ is the enthalpy in minus the enthalpy out due to mass transfer.
Consider boiling of water on a stove; water open to the atmosphere with heat added to the water. At steady state, the temperature of the water is constant, as you state. Consider the system as the water alone; this is an open system with mass out due to evaporation.  If the evaporation takes place at constant pressure (infinite surroundings) the temperature of the water remains at the saturation temperature at that pressure, a constant temperature.  The internal energy of the water decreases due to loss of water as vapor and that decrease in internal energy of the water is due to loss of mass of water, not a decrease in temperature. (The specific internal energy, internal energy per unit mass, does not change, but the total internal energy of the mass of water decreases.)
A pressure cooker is a different situation. Heat added increases both pressure and temperature.  This case can be evaluated as a closed system (water and air/vapor in the vessel) until the pressure relief valve opens, then the mass loss out the valve must be considered by treating the system as an open system.
Additional information
Evaporation (and Condensation) are mass transfer processes driven by pressure differences between the saturation pressure of liquid at its temperature, and the partial pressure of the liquid's vapor in the surroundings (humidity).  There can also be heat transfer between the liquid and its surroundings driven by temperature difference.
Evaporation is the process of turning liquid into vapor. (Condensation is the process of turning vapor into liquid.)
Typically, evaporation and boiling are evaluated with air in the surroundings, and pressure of the surroundings is the air pressure plus the vapor pressure.
Evaporation is called Boiling when the liquid temperature is at the saturation temperature for the pressure of the surroundings, and the vapor pressure in the surroundings is less than the saturation pressure of the liquid.
For the liquid temperature lower than the saturation temperature at the pressure of the surroundings, evaporation still occurs but at a much lower rate.
If a liquid is suddenly exposed to pressure lower than the saturation pressure at the liquid's temperature, with no air present, the very rapid ("instantaneous") evaporation is called Flashing.  This occurs in a heat pump refrigeration cycle when the liquid passes through the expansion (throttling) valve into the evaporator.
Another interesting process is adiabatic saturation, used in swamp coolers in dry parts of the country, such as New Mexico.  Also, the sling psychrometer is an interesting process.
A: Temperature is the manifestation of the kinetic energy of molecules.
Here's four clarifications you need to have.

*

*Temperature is directly proportional to the kinetic energy of the molecules of a liquid.

*Internal energy = kinetic energy + potential energy.

*Potential energy has very, very little to do with temperature, and depends on the distance between the molecules.

*Molecules in a gas are farther apart from each other compared to liquids.

To boil a liquid, that is, convert it from liquid to gas, the heat supplied increases the distance between the molecules, thereby increasing its potential energy. Since there is a change in potential energy, there is a change in internal energy. But since kinetic energy remains constant, there is no change in temperature.
Remember, heat energy is used for two things - to make a liquid hotter (i.e., increase the temperature / kinetic energy) + to change the state of a liquid at boiling point (i.e., to increase the potential energy, not temperature).
The energy supplied to change the state of a system (i.e., potential energy) is called latent heat. So, supplying heat energy always changes internal energy but it need not change temperature.
Perhaps, the reason why you had this confusion is because in an ideal gas, all internal energy is kinetic (potential energy = 0). Although this assumption is made to simplify certain ideas in Physics, we cannot ignore potential energy here.
A: The water molecules have an average distance as a function of temperature. From $4^{\circ}$C to $100^{\circ}$C, the effect of heat is to increase that distance. The energy of molecules, as the case in a system mass-spring, is a sum of a potential and a kinetic term. They vibrate around the average distance. So the effect of the increase in the temperature is not only to increase the kinetic energy.
In the boiling point, the molecular bonds are broken, and all the energy is kinetic in the gas phase. But it is required energy to break that bonds, and the heat input is the source of that energy.
The inverse happens when water vapour condenses to liquid. Energy is delivered by the system to the surroundings in the process when the molecular bonds are made.
So, that heat that goes in and out of the system according to the direction of the phase change is the variation of its internal energy.
A: Contrary to the premise of the question, and contrary to several existing answers, the relationship between temperature and molecular kinetic energy is a result, not an axiom or an assumption.  Fundamentally, temperature is a relationship between internal energy and entropy:
\begin{align}
\mathrm dU &= T\,\mathrm dS + \cdots \\
\frac{\partial S}{\partial U} &= \frac 1T
\end{align}
To un-math: imagine two systems where the same increase in internal energy, $\mathrm dU$, results in different increases in entropy, $\mathrm dS$.  If some fixed amount of energy moves from the system with small ${\partial S_1}/{\partial U}$ to the system with large ${\partial S_2}/{\partial U}$, the small decrease in the first entropy is offset by a larger increase in the second entropy.  Entropy-increasing exchanges of energy happen spontaneously, so the first system will tend to give energy to the second, but not the other way around. We call the generous system “hot” and the greedy system “cold.”
In statistical mechanics there are four useful variations of the “internal energy.”

*

*The traditional internal energy $U$ depends only on the intrinsic properties of the material.  As a rule, changes $\mathrm dU$ are observable, but the “total energy content” $U$ is not.


*Changes in the “Helmholtz free energy,” $\mathrm dA=\mathrm dU-T\mathrm dS$, measure the reversible work that can be extracted at constant temperature.


*A change in the “enthalpy,” $\mathrm dH=\mathrm dU+p\mathrm dV$, accounts for both change in internal energy and also the work done on the environment by a volume change $\mathrm dV$, which depends on the pressure $p$.


*A change in the “Gibbs free energy,” $\mathrm dG=\mathrm dU+p\mathrm dV-T\mathrm dS$, is the maximum non-expansion work that can be extracted from a system.
You ask about boiling.  For liquid water turning to vapor in an open pan, the energy required is the enthalpy of vaporization. The enthalpy of vaporization varies with temperature, and has the same magnitude whether you consider boiling or evaporation.
The crux of your question is here:

I have been taught that temperature represents the kinetic energies of the molecules of the system that is internal energy. In this case, the internal energy is clearly increasing, since we are transitioning from liquid to vapour phase, then why doesn't the temperature change?

The problem is that what you learned was wrong — right enough to be useful in many circumstances, but wrong enough to trip you up here.  Instead of thinking of temperature as energy content, think of temperature as energy generosity.  For the most part, energy generosity is proportional to energy content within a particular phase; however, phase changes are complicated.
You already have an answer which discusses phase changes as a mechanism for a system to reduce its (Helmholtz or Gibbs) free energy.  In a nutshell, you get vaporization when some “specific” amount of your material can increase its entropy by changing phases — so long as the enthalpy required for the phase transition is available.
For boiling and sublimation, especially at low pressures, I find it more useful to think in terms of vapor pressures; see this related question, and links therein.
If the crux of your question is whether the temperature during a phase transition is really constant, you can think of the phase transition as a terrifically efficient mechanism for bringing a superheated or supercooled system back to its phase-equilibrium temperature for the ambient pressure. Think of cooking on a stove at a low simmer, and spilling a little water on your hot stove element, where the water boils like gangbusters — perhaps even exhibiting the Leidenfrost effect, where boiling is slowed by a constantly-replenished vapor layer between the liquid and some superheated surface. You might also think of those supercooled sodium acetate heating pads, which suddenly warm up to about 50°C when you trigger them to solidify.
Perhaps another way to answer your question, going back to the definition at the top: steam generates more entropy per unit of internal energy than liquid water does.  If you tried to raise liquid water above its boiling point, it would generate more entropy to convert the water to steam and store the energy in the steam instead.  So the phase change is what happens. Because  the phase change is more efficient at removing heat from the liquid than your heat source is at adding heat, the liquid water temperature gets “latched” at the boiling point.
A: What happens when water boils and why does the temperature remain constant despite adding additional heat to the liquid?
Intermolecular forces provide the cohesion that separates a liquid from a vapor. The energy required to overcome these forces and change the liquid to a vapor is the internal energy U. While the kinetic energy of the liquid molecules is less than the binding energy (U), the temperature in the liquid will rise as heat is added since all this heat will go into work that increases the kinetic energy of the liquid molecules. This process occurs until the boiling point is reached. At the boiling point, the molecules in the liquid have reached their maximum kinetic energy. Any additional heat goes into weakening the interatomic forces until they disappear evetually freeing the molecule from the liquid into the gas with no change in kinetic energy.
There is also work done by the heat provided in the expansion of the vapor against the ambient pressure. Since boiling is an isobaric process, the work due to expansion is simply $P\Delta V$
All this allows for an expression for the heat required for vaporization.
$Q_v = U + P\Delta V$
Note that this expression does not include the heat required to raise the temperature of the liquid to its boiling point. Only the heat required to overcome the intermolecular forces and allow for gas expansion. Until the water is completely evaporated, none of the heat will increase the kinetic energy of the vapor so the temperature remains constant until all the liquid is gone.
