50g of ice at 0 degrees C mixed with 200g water at 40 degrees C — find final temperature

Question

50 g of ice at 0°C is mixed with 200 g of water at 40°C. Find the final temperature of the mixture. ( $L_f = 80$ cal/g, $c_w = 1$ cal/g·°C)

Solution — Step by Step

Before setting up the equilibrium equation, we must check whether the hot water has enough heat to melt all the ice completely.

Heat available from 200 g of water cooling from 40°C to 0°C:

Q_{available} = mc\Delta T = 200 \times 1 \times 40 = 8000 \text{ cal}

Heat needed to melt all 50 g of ice:

Q_{melt} = mL_f = 50 \times 80 = 4000 \text{ cal}

Since $Q_{available} = 8000$ cal $> Q_{melt} = 4000$ cal, all the ice melts and there is still heat remaining in the water.

Remaining heat after melting: $8000 - 4000 = 4000$ cal.

This remaining 4000 cal will now heat the 250 g of liquid water (200 g original + 50 g from melted ice) — all starting at 0°C.

Q_{remaining} = m_{total} \times c_w \times T_f

4000 = 250 \times 1 \times T_f

T_f = \frac{4000}{250} = 16°C

Heat lost by hot water = Heat used to melt ice + Heat to warm all water

$200 \times 1 \times (40 - 16) = 50 \times 80 + 250 \times 1 \times 16$

$200 \times 24 = 4000 + 4000$

$4800 = 8000$ — wait, this doesn’t balance!

Let me redo: Heat lost by 200 g water going from 40°C to $T_f$ :

Q_{lost} = 200 \times 1 \times (40 - T_f)

Heat gained: melt ice + warm 250 g water from 0°C to $T_f$ :

Q_{gained} = 50 \times 80 + 250 \times 1 \times T_f

Setting equal:

200(40 - T_f) = 4000 + 250 T_f

8000 - 200 T_f = 4000 + 250 T_f

4000 = 450 T_f

T_f = \frac{4000}{450} \approx 8.89°C \approx \mathbf{8.9°C}

Why This Works

The principle of calorimetry says heat lost by the hot body equals heat gained by the cold body (in an insulated system). Here, the cold body (ice) goes through two stages: phase change (melting) at constant temperature, then temperature rise.

We must always check whether the phase change completes before assuming a final temperature above 0°C.

Alternative Method — Quick Check

The two-step method in my earlier steps was slightly off (I incorrectly assumed the melted ice and original water are treated separately). The correct unified approach is the energy balance in Step 4, giving $T_f \approx 8.9°C$ .

The systematic energy balance is always more reliable than the two-step approach for this type of problem.

Common Mistake

The classic error: forgetting that after the ice melts, we now have 250 g of cold water (not 200 g) that needs to be heated to the final temperature. Students use 200 g in the denominator and get a wrong final temperature. After ice melts, all the water — original hot water plus water from melted ice — participates in reaching thermal equilibrium.

Question

Solution — Step by Step

Why This Works

Alternative Method — Quick Check

Common Mistake

Try These Next