Lets say I have ice of mass $m_i$ and initial temperature $T_i$ and specific heat $s_i$ . And I have water of mass $m_w$ and initial temperature $T_w$ and specific heat $s_w$ . I have put both this water and ice in a copper container with mass $m_{cu}$ , specific heat $s_{cu}$ and initial temperature $T_w$ . What will be the final temperature $T$ of the system ? ( The Latent heat of freezing of water is $L_f$ )

Here ,
$$Q_1 = m_is_i(T-T_i) + m_iL_f$$ $$Q_2 = m_ws_w(T-T_w) + m_wL_f + m_{cu}s_{cu}(T-T_w)$$

According to Calorimetry , $$Q_1 = Q_2$$ $$m_is_i(T-T_i) + m_iL_f = Q_2 = m_ws_w(T-T_w) + m_wL_f + m_{cu}s_{cu}(T-T_w)$$

I can use this equation to calculate $T$ , But then I ran into a problem. In this case , I don't know which is happening , the freezing of water , or the melting of ice , so i can't just put both $m_wL_f$ and $m_iL_f$ in the equation and call it done. I might have to remove one of them depending on which thing is not happening (freezing or melting) .To make matters worse , The water or ice may just freeze or melt partially. Is there another equation that can solve for the final temperature while also taking all of this into account?

  • $\begingroup$ You can assume the ice will melt and that the water volume will increase. $\endgroup$ – Yejus Nov 23 '20 at 7:14
  • $\begingroup$ All components will end up at the same temperature. This is NOT NECESSARILY $T>0$.. Your expression of $Q_1$ is wrong because once melted the water heat up further. $\endgroup$ – Gert Nov 23 '20 at 9:57

You actually have to consider all 3 cases: all the ice melts, all the water freezes, or you end up with a mixture of ice and water. Also, it is much more convenient to work with the first law of thermodynamics in solving this. Because we are assuming that this system is isolated, it tells us that the change in internal energy of the system is zero.

Let $m_{wi}$ and $m_{ii}$ be the masses of water and ice initially and $m_{wf}$ and $m_{if}$ be the masses of water and ice finally. Also, let the reference state for internal energy be water at 0 C. Then the specific internal energy of the water and ice initially are $$u_{wi}=s_w(T_w-0)$$ $$u_{ii}=-L_f-s_i(0-T_i)$$and the specific internal energy of water and ice finally are $$u_{wf}=s_w(T-0)$$ $$u_{if}=-L_f-s_i(0-T)$$So setting the overall internal energy of the system initially to the overall internal energy of the system finally gives: $$m_{wi}s_wT_{w}+m_{ii}(-L_f+s_iT_i)+m_{cu}s_{cu}T_w=m_{wf}s_wT+m_{if}(-L_f+s_iT)+m_{cu}s_{cu}T$$ This equation is subject to the constraint that the total mass of water and ice is constant:$$m_{wi}+m_{ii}=m_{wf}+m_{if}$$

The three possible cases are then as follows:

  1. If water and ice are both present finally, then T = 0
  2. If only water is present finally, then $m_{if}=0$
  3. If only ice is present finally, then $m_{wf}=0$
  • $\begingroup$ Is there a mathematical way to check if all the water has melted or all the ice has melted or if both water and ice are present ?(I am asking this because I want it to be consistent with the rest of my understanding) . Also thanks for the idea of "constraint mass". $\endgroup$ – Prithu biswas Nov 23 '20 at 14:45
  • $\begingroup$ If you look at the results from the three cases, the results for two of the cases won't make sense. For example, if case 1 is actually incorrect, and the final temperature is not 0, then the final mass of ice or of water will be calculated to be negative if T is set to 0 . $\endgroup$ – Chet Miller Nov 23 '20 at 14:54

Assume the end-temperature to be $T$. Your expression for $Q_1$ is then wrong.

The ice will heat from $T_i$ to $0$, then absorb latent heat of melting, then the melted water will heat to $T$, so we have:


  • $\begingroup$ But what will happen if the ice is so cold and the water is so little amount that the ice will end up freezing all the water in the copper container? $\endgroup$ – Prithu biswas Nov 23 '20 at 11:47
  • $\begingroup$ You can figure that out by trial and error. First assume all ice melts and calculate $T$. If $T>0$ you're right. If not, try $T=0$. If that doesn't fit try the solution you propose: everything freezes at a temperature $T>T_i$. $\endgroup$ – Gert Nov 23 '20 at 11:59

As others have correctly answered, there are a number of different mathematically possible solutions.

This leads to the possibility of trying and discarding a number of solutions before finding the correct one.

A useful technique (I think) for avoiding wasted effort is as follows:

  1. Pick an arbitrary guess as to the final state; say, all the ice warms to $0^o$C and melts, and the water and the copper all cool to $0^o$C. This guess is not unique; it does lead to simpler calculations, but other possible guesses will work just as well. In fact, this particular guess is very unlikely to be correct.
  2. Calculate the energy to be added/removed to achieve this guess state. For example, you need to add energy to warm the ice, and more energy to melt it. You need to remove energy to cool the water to $0^o$C, and more energy to cool the copper container to $0^o$C. All of these can be calculated.
  3. Now combine these different energy flows to find the net energy flow into/out of the system.

If the energy flow is already zero, then your original guess was actually correct. Job done.

if you have a net energy output, then you need to put back this amount to achieve zero net energy. You did this in an insulated calorimeter, right?

The added energy will warm the original water, the melted ice water and copper from their common guess value to a final state. Unless the system gets heated to above boiling (very hot original copper), you're done.

If Step #$3$ gives you a net energy input, then you need to remove this amount to achieve zero net energy. This removal of heat will freeze a calculated amount of the water. If all the water freezes and you still need to remove heat, then the copper and ice will cool together to some common, sub-zero temperature. Job done


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