A glass of cold lassi at a Punjabi dhaba often comes with a heap of crushed ice. Sometimes the ice melts away in a few minutes. Sometimes a lump of ice floats in the glass long after the lassi has turned icy cold. The answer depends on two numbers, not on patience.
A tempting mistake
Put \(100\,\text{g}\) of ice at \(0\,\text{°C}\) into \(200\,\text{g}\) of water at \(30\,\text{°C}\) in an insulated vessel. Suppose all the ice melts and the mixture ends at \(T\). Heat gained: \(0.1 \times 3.34 \times 10^5 + 0.1 \times 4200 \times T\). Heat lost: \(0.2 \times 4200 \times (30 - T)\). Solving gives \(T \approx -6.5\,\text{°C}\). Water colder than 0 °C, with ice melting in it, is impossible. The assumption that all the ice melts was wrong.
So first compare two heats. The first is the heat the warm water can give if it cools all the way to 0 °C: \(Q_\text{avail} = m_w c_w T_w\). The second is the heat needed to melt all the ice: \(Q_\text{melt} = m_i L_f\). This lesson uses \(c_w = 4200\), \(c_\text{ice} = 2100\) and \(c_\text{steam} = 2010\,\text{J/(kg K)}\), with \(L_f = 3.34 \times 10^5\) and \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
Let \(Q_\text{avail}\) be the heat the warm body can give in cooling to 0 °C. Let \(Q_\text{melt}\) be the heat to bring all the ice to 0 °C and melt it. If \(Q_\text{avail} < Q_\text{melt}\), the mixture ends at 0 °C with some ice left. If \(Q_\text{avail} = Q_\text{melt}\), all the ice just melts and the mixture is water at 0 °C. If \(Q_\text{avail} > Q_\text{melt}\), all the ice melts and the final temperature is above 0 °C.
For the example, \(Q_\text{avail} = 0.2 \times 4200 \times 30 = 25\,200\,\text{J}\) and \(Q_\text{melt} = 0.1 \times 3.34 \times 10^5 = 33\,400\,\text{J}\). The water runs out of heat first. It cools to 0 °C, and its 25 200 J melts \(25\,200/3.34 \times 10^5 = 0.0754\,\text{kg}\) of ice. So the vessel ends at 0 °C with 75.4 g of the ice melted and 24.6 g left. It holds 275.4 g of water.
Try the sliders in the figure above. With 100 g of ice, the bars cross when the water is at about 40 °C: at \(0.1 \times 3.34 \times 10^5/(0.2 \times 4200) = 39.8\,\text{°C}\). Below that the box reads 0 °C with some ice left. Above it all the ice melts and the final temperature climbs above 0 °C. Exactly at the crossing the ice just melts and everything is water at 0 °C.
\(50\,\text{g}\) of ice at \(0\,\text{°C}\) is put into \(50\,\text{g}\) of water at \(40\,\text{°C}\) in an insulated vessel. What mass of ice is left when the mixture settles, in grams? Take \(c_w = 4200\,\text{J/(kg K)}\) and \(L_f = 3.34 \times 10^5\,\text{J/kg}\).
\(100\,\text{g}\) of ice at \(0\,\text{°C}\) is added to \(200\,\text{g}\) of water at \(40\,\text{°C}\). Take \(c_w = 4200\,\text{J/(kg K)}\) and \(L_f = 3.34 \times 10^5\,\text{J/kg}\). What is in the vessel at the end?
When all the ice melts
When \(Q_\text{avail} > Q_\text{melt}\), the ice goes through every step to the final temperature \(T\). If it starts below 0 °C, it first warms to 0 °C. Then it melts. Then the melted water warms to \(T\). The warm water gives all of this as it cools to \(T\). Put the steps on one side and the loss on the other.
\(50\,\text{g}\) of ice at \(-10\,\text{°C}\) is dropped into \(200\,\text{g}\) of water at \(40\,\text{°C}\) in an insulated flask. Find the final temperature.
Ice needs: warm to 0 °C, \(0.05 \times 2100 \times 10 = 1050\,\text{J}\); melt, \(0.05 \times 3.34 \times 10^5 = 16\,700\,\text{J}\). Total 17 750 J. The water can give \(0.2 \times 4200 \times 40 = 33\,600\,\text{J}\).
\(17\,750 + 0.05 \times 4200 \times T = 0.2 \times 4200 \times (40 - T)\).
\(17\,750 + 210T = 33\,600 - 840T\), so \(1050T = 15\,850\) and \(T \approx 15.1\,\text{°C}\).
The mixture ends at about 15.1 °C, all of it water. The ice took more than half of the water's available heat just to melt.
Fig 5.1 shows how the final state moves as more ice is added to 100 g of water at 40 °C. The water can give \(0.1 \times 4200 \times 40 = 16\,800\,\text{J}\), which melts at most \(16\,800/334 = 50.3\,\text{g}\) of ice. Up to B, all the ice melts and the temperature falls. Beyond B, the temperature stays at 0 °C and the extra ice stays solid. With 60 g added, the vessel holds 9.7 g of ice and 150.3 g of water.
\(30\,\text{g}\) of ice at \(0\,\text{°C}\) is added to \(300\,\text{g}\) of nimbu pani at \(30\,\text{°C}\). Treat the drink as water, \(c = 4200\,\text{J/(kg K)}\), with no heat lost. Take \(L_f = 3.34 \times 10^5\,\text{J/kg}\). What is the final temperature?
\(20\,\text{g}\) of ice at \(-20\,\text{°C}\) is put into \(100\,\text{g}\) of water at \(25\,\text{°C}\). Take \(c_\text{ice} = 2100\), \(c_w = 4200\,\text{J/(kg K)}\) and \(L_f = 3.34 \times 10^5\,\text{J/kg}\). What is the final temperature?
Steam into water
Steam is the mirror image of ice. Passed into cold water, it condenses at 100 °C and gives out \(L_v\) per kilogram. Then the condensed water cools to the final temperature and gives out more. Because \(L_v\) is so large, a little steam heats a lot of water. The comparison works the same way, with 100 °C in place of 0 °C.
\(10\,\text{g}\) of steam at \(100\,\text{°C}\) is passed into \(200\,\text{g}\) of water at \(20\,\text{°C}\) in an insulated vessel, and all of it condenses. Find the final temperature.
The cold water can take \(0.2 \times 4200 \times 80 = 67\,200\,\text{J}\) before it boils. Condensing all the steam gives only \(0.01 \times 2.26 \times 10^6 = 22\,600\,\text{J}\).
\(22\,600 + 0.01 \times 4200 \times (100 - T) = 0.2 \times 4200 \times (T - 20)\).
\(22\,600 + 4200 - 42T = 840T - 16\,800\), so \(882T = 43\,600\) and \(T \approx 49.4\,\text{°C}\).
Ten grams of steam brings 200 g of water from 20 °C to about 49.4 °C. Ten grams of boiling water would raise it by less than 4 K.
Sometimes the cold side cannot take all the steam's heat. Then the mixture stops at 100 °C and some steam is left. Steam passed into water until the water just boils is the edge case. The final temperature is 100 °C, so the condensed steam does not cool at all. Each gram gives only its latent heat. Bringing 200 g of water from 30 °C to the boil needs \(0.2 \times 4200 \times 70 = 58\,800\,\text{J}\), which is \(58\,800/2260 = 26.0\,\text{g}\) of steam.
\(10\,\text{g}\) of steam at \(100\,\text{°C}\) is passed into \(50\,\text{g}\) of ice at \(0\,\text{°C}\) in an insulated vessel. All the ice melts. What is the final temperature? Take \(c_w = 4200\,\text{J/(kg K)}\), \(L_f = 3.34 \times 10^5\) and \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
Steam at \(100\,\text{°C}\) is passed into \(100\,\text{g}\) of ice at \(0\,\text{°C}\) in an insulated vessel. What mass of steam just melts all the ice, so that everything ends as water at \(0\,\text{°C}\)? Use the values of this lesson.
\(5\,\text{g}\) of steam at \(100\,\text{°C}\) is passed into \(100\,\text{g}\) of ice at \(0\,\text{°C}\). Which statements are true? Use the values of this lesson.
A mixture of ice and steam has three possible final states. It may end at 0 °C with ice left, between 0 °C and 100 °C, or at 100 °C with steam left. Test the two edges before writing an equation. Find the heat to bring everything to 0 °C, and the heat to bring everything to 100 °C. The final state lies where the heat balance changes sign.
Mix \(1\,\text{kg}\) of ice at \(-20\,\text{°C}\) with \(1\,\text{kg}\) of steam at \(200\,\text{°C}\) (steam kept at 1 atm). Bringing the ice to water at 100 °C needs 42 000 J to warm the ice, 334 000 J to melt it and 420 000 J to warm the water. That is 796 000 J in all. Cooling the steam to 100 °C gives only \(1 \times 2010 \times 100 = 201\,000\,\text{J}\). The other 595 000 J must come from condensing steam: \(595\,000/(2.26 \times 10^6)\), or 0.263 kg. So the mixture ends at 100 °C, with 1.263 kg of water and 0.737 kg of steam.
Now the lassi at the dhaba. A 300 g glass at 30 °C can give about \(0.3 \times 4200 \times 30 = 37\,800\,\text{J}\) in cooling to 0 °C. That melts at most \(37\,800/334 \approx 113\,\text{g}\) of ice. Add less, and all of it melts in the end. Add a heap of 150 g, and about 37 g of ice stays floating in the ice-cold glass. It stays as long as no heat reaches the glass. Patience changes nothing; the comparison of heats decides.
- Solving for T without first comparing the heats.Compare the heat available in cooling to 0 °C with the heat needed to melt all the ice. A negative answer means the wrong case.
- Forgetting to warm ice from below 0 °C before melting it.Ice at −10 °C first takes \(m c_\text{ice} \times 10\), with \(c_\text{ice} = 2100\), then \(mL_f\).
- Forgetting that the melted ice, or the condensed steam, must still change temperature.After melting, the melt warms from 0 °C to T; after condensing, the condensate cools from 100 °C to T.
- Treating the final state as ice at a temperature below 0 °C in a vessel of warm water.If ice and water are both present at the end, the temperature is exactly 0 °C.