## Mixing Ice and Water of Two Different Temperatures

### Task number: 1280

An ice cube with a temperature of −10 °C and a mass of 2 kg is tossed into water with a temperature of 70 °C and a mass of 1 kg. Immediately after that more water with a temperature of 40 °C and a mass of 1 kg is poured in.

The entire process proceeds under the normal atmospheric pressure.

Determine the state, on which the system stabilizes after reaching thermodynamic equilibrium.

• #### Hint

What happens with the temperature and heat when we immerse a cold body into a hot water?

• #### Analysis

Since the water has a higher temperature than the ice at the beginning of this process, the heat will be transferred from water to ice. This happens because the heat is transferred from a warmer to a colder body. The heat supplied to the ice is consumed to raise the temperature of the ice to the melting point, then to melt the ice and eventually (if the hot water supplies enough heat) to warm up the melted water to the resulting temperature.

The final state of the system depends on how much heat the water supplies. If the amount of supplied heat is larger than the heat necessary for melting all the ice, then the resulting state will be liquid water with a temperature above 0 °C. The supplied heat is used for melting the ice and the rest is used for warming the melted water.

Furthermore, the supplied heat can equal exactly the heat needed for melting the ice, creating water at a temperature of 0 °C.

Another possibility is that the supplied heat is not hot enough to melt the ice completely. In this case, a mixture of water and ice is created with a temperature of 0 °C. The amount of the melted ice depends on the difference between the supplied heat and the heat used for warming the ice to the melting point.

The last possibility is that the heat supplied by the water is not hot enough to raise the temperature of the ice to the melting point. Then, part of the water freezes and the result is again a mixture of water and ice with a temperature of 0 °. Or else, all of the water freezes which results in a solid state (ice).

When solving this task, it is necessary to compare the values of each heat and decide which case will occur.

• #### Numerical values

 m1 = 1 kg the initial mass of water t1 = 70 °C the initial water temperature m2 = 2 kg the mass of added ice t2 = −10 °C the temperature of added ice m3 = 1 kg the mass of added water t3 = 40 °C the temperature of added water tm = 0 °C the melting point of ice mm = ? the mass of melted ice

From the Handbook of chemistry and physics:

 cw = 4180 J kg−1 K−1 the specific heat capacity of water ci = 2100 J kg−1 K−1 the specific heat capacity of ice lfus = 334 kJ kg−1 = 334·103 J kg−1 the specific latent heat of fusion of ice
• #### Solution

The most important thing is to find out whether the water is hot enough (and of sufficient amount) to supply the heat needed to melt the ice. For this purpose, we calculate the sum of heats Q1r and Q2r released by the two masses of water with different temperatures when they cool down to the melting point tm = 0 °C.

The heats are:

Q1r = m1cw(t1tm),

Q2r = m3cw(t3tm),

Q1r + Q2r = m1cw(t1tm) + m3cw(t3tm),

Q1r + Q2r = [1 · 4180 · (70 − 0) + 1 · 4180 · (40 − 0)] J

= 459.8·103 J = 459.8 kJ .

Now we determine the amount of the heat needed to melt all of the ice. We have to realize that it is the sum of the heat Q1s needed to raise the temperature of 2 kg of ice to the melting point and the heat Q2s needed to melt the ice.

Therefore:

Q1s = m2ci(tmt2),

Q2s = lfusm2,

Q1s + Q2s = m2ci(tmt2) + lfusm2,

Q1s + Q2s = [2 · 2100 · (0 − (−10)) + 334·103 · 2] J =

= (42·103 + 668·103 ) J = 710 kJ.

We can see that the heat supplied by the water when it cools down to the melting point of ice is quite sufficient to heat up the ice to the melting point, but it is not sufficient enough to melt the ice cube completely. The resulting state of the system will therefore be such that the water cools down to 0 °C, the ice heats up to 0 °C and a certain part of the ice with the mass mm melts. We determine this mass by the following considerations:

The heat supplied when the water cools down is Q1r+ Q2r. Now we need to subtract the heat Q1s needed to raise the ice temperature to the melting point. The remainder is used up to melt the ice.

Therefore:

mmlfus = Q1r + Q2rQ1s .

We evaluate mm from the last equation:

$m_{m} = \frac{Q_{1r} + Q_{2r} - Q_{1s}}{l_{fus}}\,.$
• #### Numerical insertion

$m_{m} = \frac{Q_{1r} + Q_{2r} - Q_{1s}}{l_{fus} }$ $m_{m} =\frac{459{.}8\cdot{10^{3}} - 2\cdot{ 2100}\cdot \left( 0-\left( -10\right)\right)}{334\cdot {10^{3}}}\, \mathrm{kg}\dot{=} 1{.}25\, \mathrm{kg}$  