Five legs from ice to steam
Take \(100\,\text{g}\) of ice at \(-20\,\text{°C}\) in a covered beaker. A heater gives it a steady \(1000\,\text{W}\), and no heat is lost. Follow its temperature until it is steam at \(120\,\text{°C}\). The two lessons before give every step. This lesson uses \(c_\text{ice} = 2100\), \(c_\text{water} = 4200\) and \(c_\text{steam} = 2010\,\text{J/(kg K)}\), with \(L_f = 3.34 \times 10^5\) and \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
Before you play the figure below, predict: which stage of this run lasts longest?
Read Fig 3.1 leg by leg. Each leg's heat is \(Q = mc\,\Delta T\) on a slope and \(Q = mL\) on a plateau. At 1000 W the time in seconds is the heat in kilojoules.
- Warm the ice from −20 °C to 0 °C: \(0.1 \times 2100 \times 20 = 4200\,\text{J}\), which takes 4.2 s.
- Melt the ice at 0 °C: \(0.1 \times 3.34 \times 10^5 = 33\,400\,\text{J}\), 33.4 s.
- Warm the water from 0 °C to 100 °C: \(0.1 \times 4200 \times 100 = 42\,000\,\text{J}\), 42 s.
- Boil the water at 100 °C: \(0.1 \times 2.26 \times 10^6 = 226\,000\,\text{J}\), 226 s.
- Warm the steam from 100 °C to 120 °C: \(0.1 \times 2010 \times 20 = 4020\,\text{J}\), about 4.0 s.
The whole run takes about 310 kJ and 310 s. The boiling plateau is \(226/42 \approx 5.4\) times as long as heating the water through 100 K.
A heating curve is a graph of the temperature of a substance against time, while heat is supplied at a steady rate. It may also be drawn against the heat supplied. Sloping parts show one state warming; flat parts show a change of state at a fixed temperature.
How much heat turns \(0.5\,\text{kg}\) of ice at \(-10\,\text{°C}\) into steam at \(110\,\text{°C}\)? What share of it goes into boiling? Use the values of this lesson.
Five steps: warm the ice by 10 K, melt it, warm the water by 100 K, boil it, warm the steam by 10 K.
\(Q = m c_\text{ice}(10) + m L_f + m c_\text{water}(100) + m L_v + m c_\text{steam}(10)\) with \(m = 0.5\,\text{kg}\).
The five steps take 10 500 J, 167 000 J, 210 000 J, 1 130 000 J and 10 050 J. The total is 1 527 550 J, about 1.53 MJ.
About 1.53 MJ is needed, and boiling takes 1.13 MJ of it, 74 %.
How much heat turns \(1\,\text{kg}\) of ice at \(-10\,\text{°C}\) into water at \(20\,\text{°C}\), in kilojoules? Take \(c_\text{ice} = 2100\), \(c_\text{water} = 4200\,\text{J/(kg K)}\) and \(L_f = 3.34 \times 10^5\,\text{J/kg}\).
What a slope says, what a plateau says
On a sloping leg the heater supplies \(P\,\Delta t\) in a time \(\Delta t\), and this equals \(mc\,\Delta T\). So the slope of the graph is fixed by the heater and the substance. A steep slope means a small \(mc\). This is why the ice leg of Fig 3.1 is steeper than the water leg. Ice has half the specific heat of water, so it warms twice as fast.
\[ Q = P\,\Delta t \]
Why. a steady heater of power \(P\) supplies energy \(P\) every second.
\[ P\,\Delta t = m c\,\Delta T \]
Why. on a slope all of that heat warms one state.
\[ \dfrac{\Delta T}{\Delta t} = \dfrac{P}{mc} \]
Why. divide both sides by \(mc\,\Delta t\); this is the slope.
\[ P\,\Delta t_\text{plateau} = m L \;\Rightarrow\; \Delta t_\text{plateau} = \dfrac{mL}{P} \]
Read it. on a plateau all the heat changes the state, and the temperature does not move.
Turn the rules around and a graph becomes a measurement. If you know \(P\) and \(m\), each slope gives a specific heat, \(c = P\,\Delta t/(m\,\Delta T)\). Each plateau gives a latent heat, \(L = P\,\Delta t/m\). The two plateaus of water, on one heater, have lengths in the ratio \(L_v/L_f = 2.26 \times 10^6/3.34 \times 10^5 \approx 6.8\).
Fig 3.2 shows the heating curve of \(0.25\,\text{kg}\) of a solid on a \(200\,\text{W}\) heater, with no heat lost. Find its melting point, its specific heat as a solid and as a liquid, and its latent heat of fusion.
The flat part AB is at 80 °C, so that is the melting point. OA is the solid warming, AB the melting, BC the liquid warming.
On a slope \(c = P\,\Delta t/(m\,\Delta T)\); on the plateau \(L = P\,\Delta t/m\).
Solid: \(200 \times 60/(0.25 \times 60) = 800\,\text{J/(kg K)}\). Melting: \(200 \times 150/0.25 = 1.2 \times 10^5\,\text{J/kg}\). Liquid: \(200 \times 80/(0.25 \times 40) = 1600\,\text{J/(kg K)}\).
Melting point 80 °C; \(c = 800\,\text{J/(kg K)}\) as a solid and \(1600\,\text{J/(kg K)}\) as a liquid; \(L_f = 1.2 \times 10^5\,\text{J/kg}\).
A \(1.5\,\text{kW}\) kettle holds \(1.2\,\text{kg}\) of water that has just reached \(100\,\text{°C}\). How much longer does it take to boil away \(0.3\,\text{kg}\) of the water, in seconds? Neglect losses. Take \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
A graph of temperature against the heat supplied, for \(1\,\text{kg}\) of a substance, rises from 0 °C to 50 °C over the first 50 kJ. It stays at 50 °C from 50 kJ to 150 kJ, then rises again. Which statements are true?
Cooling curves
Run the film backwards. A test tube of molten naphthalene is left in a room at 20 °C. Its temperature falls, then stays flat at about 80 °C, then falls again. This is a cooling curve. The flat part is the freezing point, the same temperature as the melting point. There the liquid gives out its latent heat as it solidifies. That heat exactly replaces the heat lost to the room, so the temperature holds still.
Molten wax cools in a room. Its temperature falls, stays at \(57\,\text{°C}\) for several minutes, then falls again. What is happening during the flat part?
Two substances on one graph
Heat equal masses of two solids, A and B, on identical heaters and draw both curves on one graph. Three questions can be answered by eye. The higher plateau belongs to the higher melting point. The steeper solid slope belongs to the smaller specific heat, since the slope is \(P/(mc)\) with the same \(P\) and \(m\). The longer plateau belongs to the larger latent heat, since its length is \(mL/P\).
Use Fig 3.3. Find the melting points of A and B, the ratio of their specific heats as solids, and the ratio of their latent heats of fusion.
Melting points: the plateaus are at 60 °C for A and 40 °C for B.
Specific heats: in the same time each receives the same heat, so heat \(= R\,t\) with a common rate \(R\). A: \(R \times 2 = m c_A \times 60\). B: \(R \times 4 = m c_B \times 40\). Dividing, \(c_A/c_B = (2/60)/(4/40) = 1/3\).
Latent heats: A's plateau lasts 4 min and B's lasts 3 min, so \(mL_A = 4R\) and \(mL_B = 3R\), and \(L_A/L_B = 4/3\).
So A melts at a higher temperature, warms three times as fast as a solid, and needs a third more heat to melt.
Equal masses of two solids, X and Y, are heated on identical heaters from the same temperature. As solids, X warms by \(30\,\text{K}\) in \(3\,\text{min}\) and Y by \(30\,\text{K}\) in \(1\,\text{min}\). Which statement is right?
Ice at \(0\,\text{°C}\) is heated at a steady rate until all of it has boiled away. How many times longer is the boiling plateau than the melting plateau? Take \(L_f = 3.34 \times 10^5\) and \(L_v = 2.26 \times 10^6\,\text{J/kg}\).
A cook who wants food at exactly 100 °C needs no thermometer: boiling water holds itself there. However high the flame, the extra heat only boils the water away faster, along the plateau of the heating curve. This is why eggs, dal and rice cook the same way in every kitchen at sea level. Turning the flame up after the water boils wastes gas and cooks nothing faster.
- Reading a steep slope as a large specific heat.The slope is \(P/(mc)\): steep means small \(c\).
- Using the whole mass when only part of it changes state.In \(Q = mL\), \(m\) is the mass that melts or boils, not the mass present.
- Reading the latent heat from the heat at the end of the plateau.The latent heat comes from the width of the plateau, end minus start.
- Using the specific heat of water for the ice or the steam legs.Each state has its own specific heat: ice 2100, water 4200, steam 2010 J/(kg K).