I am supposed to answer this question in terms of the Zeroth and First Law of Thermodynamics. So, I wrote this:

“If heat is supplied to a system, the system will absorb the heat first. In this case, the system i.e. both water and ice will absorb the heat simultaneously. However, since both water and ice are at 0° C, according to the Zeroth Law of Thermodynamics, there is not going to be any change in temperature rather the heat absorbed by the system will be used as heat of fusion to transform ice to water. This further conserves the First Law of Thermodynamics as the heat absorbed is used to do work on the ice thus increasing its internal energy.”

However, I am still unsure of the accuracy of my explanation, and that is why I am here.

  • 1
    $\begingroup$ It is certainly possible, if the system is not at equilibrium during the heat addition, to heat different parts of the water to different temperatures above 0 C. But, after adding the heat, if not all the ice melts, all the water and ice will again be at 0 C, although some of the ice will have melted. $\endgroup$ Commented May 13, 2021 at 21:56
  • $\begingroup$ Have you considered the difference in density between liquid and frozen water? $\endgroup$ Commented May 13, 2021 at 22:57

2 Answers 2


If heat is supplied to this system, it is impossible to tell which component (water or ice) will initially absorb the heat. That depends on how the heat was delivered to the system.

However as the system equilibrates, the heat will spread throughout the entire system. Any heat delivered to the ice will first go towards its latent heat of fusion. When the system equilibrates, if there is any ice remaining, then the entire system must still be at $0^{\circ}\,\textrm{C}$ by the zeroth law of thermodynamics.


It may be possible for both the water and ice to absorb the heat simultaneously if the system is heated reversibly. Consider the following thought experiment:

Let the ice and water mixture be in a perfectly insulated metal container for a long time so that we can be assured the ice and water are in thermal equilibrium with one another at 0$^o$C.

We now remove the thermal insulation from the metal container and immerse the container in a large thermal bath (constant temperature thermal reservoir) whose temperature is infinitesimally greater than 0$^o$C. This allows us to heat the ice water mixture reversibly. (If necessary, to insure uniform heating, we could include a small stirrer to circulate the water and ice, and include the stirrer work as part of the energy transfer to the mixture).

So since the thermal bath is in thermal equilibrium with the water and in thermal equilibrium with the ice, per the Zeroth law the water and ice should be in thermal equilibrium with one another.

Finally, per the first law and assuming all the ice doesn't melt, since there is no temperature increase the heat transfer from the thermal bath will equal the mass of ice in the mixture that melts times the latent heat of fusion.

Hope this helps.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.