Final temperature of a mixture of water and ice in a copper container

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:

$$Q_1=m_1s_i(T_i-0)+m_1L_f+m_ws_w(0-T)$$

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 $m_{wf}=0$

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