The Workbook

The specific heat capacity is the heat needed to warm \(1\) kg of the substance by \(1\) K: \(Q = ms\Delta T\). Its SI unit is J/kg K.

\(Q = 2 \times 4200 \times 10 = 84000\) J \(= 84\) kJ.

The heat capacity is the heat needed to warm the whole body by \(1\) K: \(C = ms\), in J/K.

\(C = 0.5 \times 400 = 200\) J/K. \(Q = C\Delta T = 200 \times 50 = 10000\) J \(= 10\) kJ.

Heat supplied \(= Pt = 100 \times 210 = 21000\) J. \(\Delta T = \dfrac{Q}{ms} = \dfrac{21000}{1 \times 4200} = 5^\circ\)C.

One mole has mass \(M\), so the molar heat capacity is \(C_m = Ms\).

Water: \(0.018 \times 4200 = 75.6\) J/mol K. Aluminium: \(0.027 \times 900 = 24.3\) J/mol K.

The water equivalent \(W\) is the mass of water that needs the same heat as the calorimeter for the same rise in temperature. So \(W s_{water} = m s\).

\(W = \dfrac{0.2 \times 400}{4200} = 0.019\) kg \(\approx 20\) g.

Treat the calorimeter as \(20\) g of extra water: \(Q = (0.1 + 0.02) \times 4200 \times 10 = 5040\) J.

1. \(A\). The same heat is supplied each second to the same mass, and \(A\) warms faster. A smaller \(s\) means less heat per degree, so a faster rise.

2. Rate of rise \(= \dfrac{P}{ms}\), so \(s \propto \dfrac{1}{\text{rate}}\). \(A\) rises \(40^\circ\)C in \(100\) s (\(0.4^\circ\)C/s) and \(B\) rises \(20^\circ\)C in \(200\) s (\(0.1^\circ\)C/s). \(s_A : s_B = 0.1 : 0.4 = 1 : 4\).

3. \(s = \dfrac{P}{m \times \text{rate}}\). \(\dfrac{s_A}{s_B} = \dfrac{m_B \times 0.1}{m_A \times 0.4} = \dfrac{2 \times 0.1}{0.4} = \dfrac{1}{2}\), so \(1 : 2\).

The latent heat is the heat needed to change the state of \(1\) kg of the substance at constant temperature. So \(Q = mL\), with \(L\) in J/kg.

During melting the heat goes into breaking the bonds that hold the molecules in the solid, not into making them move faster. So the temperature does not rise.

\(Q = 0.5 \times 3.36 \times 10^5 = 1.68 \times 10^5\) J \(= 168\) kJ.

\(Q = mL = 2 \times 2.26 \times 10^6 = 4.52 \times 10^6\) J \(= 4.52\) MJ.

Warming: \(Q = 2 \times 4200 \times 100 = 8.4 \times 10^5\) J \(= 840\) kJ. Boiling away the water takes \(\dfrac{4520}{840} \approx 5.4\) times as much heat as bringing it to the boil.

Warm the ice: \(1 \times 2100 \times 10 = 21000\) J \(= 21\) kJ.

Melt the ice: \(1 \times 3.36 \times 10^5 = 336\) kJ.

Warm the water: \(1 \times 4200 \times 20 = 84\) kJ.

Total \(= 21 + 336 + 84 = 441\) kJ.

Condensing: \(0.1 \times 2.26 \times 10^6 = 226\) kJ. Cooling: \(0.1 \times 4200 \times 50 = 21\) kJ. Total \(= 247\) kJ.

Steam first gives out its latent heat, \(226\) kJ per \(0.1\) kg, before it even begins to cool. Water at \(100^\circ\)C gives out only the \(21\) kJ of cooling. So steam delivers more than ten times the heat to the skin.

Each second the heater supplies \(336\) J. Mass melted per second \(= \dfrac{336}{3.36 \times 10^5} = 10^{-3}\) kg \(= 1\) g.

Time for \(300\) g \(= 300\) s \(= 5\) minutes.

\(Q = mL = 10^{-3} \times 2.26 \times 10^6 = 2260\) J.

\(m = \dfrac{Q}{s\Delta T} = \dfrac{2260}{4200 \times 1} = 0.538\) kg \(= 538\) g. One gram of sweat cools more than half a kilogram of the body by a degree, which is why sweating works.

1. \(PQ\) is flat at \(0^\circ\)C: the ice is melting. The flat part after \(R\) at \(100^\circ\)C: the water is boiling.

2. \(PQ\) lasts from \(10\) s to \(90\) s, that is \(80\) s. Heat \(= 420 \times 80 = 33600\) J for \(0.1\) kg. \(L = \dfrac{33600}{0.1} = 3.36 \times 10^5\) J/kg.

3. \(OP\): \(20^\circ\)C rise in \(10\) s, heat \(= 4200\) J. \(s_{ice} = \dfrac{4200}{0.1 \times 20} = 2100\) J/kg K.

   \(QR\): \(100^\circ\)C rise in \(100\) s, heat \(= 42000\) J. \(s_{water} = \dfrac{42000}{0.1 \times 100} = 4200\) J/kg K. Ice warms twice as fast as water because its specific heat is half.

1. The flat part is at \(80^\circ\)C, so naphthalene freezes (and melts) at \(80^\circ\)C.

2. From \(B\) to \(C\) the liquid is turning into solid. The latent heat released as the molecules lock into place exactly replaces the heat lost to the room, so the temperature does not fall.

3. Along \(AB\) it is all liquid, cooling. Along \(CD\) it is all solid, cooling. \(CD\) flattens out because the temperature difference from the room shrinks, so heat leaves more slowly.

Solid: heat \(= 100 \times 20 = 2000\) J for \(40^\circ\)C. \(s = \dfrac{2000}{0.1 \times 40} = 500\) J/kg K.

Melting: heat \(= 100 \times 60 = 6000\) J. \(L = \dfrac{6000}{0.1} = 60000\) J/kg \(= 60\) kJ/kg.

Liquid: heat \(= 100 \times 120 = 12000\) J for \(60^\circ\)C. \(s = \dfrac{12000}{0.1 \times 60} = 2000\) J/kg K.

1. The flat part \(PQ\) is at \(50^\circ\)C, the melting point.

2. \(PQ\) runs from \(50\) kJ to \(150\) kJ: \(100\) kJ melts \(1\) kg, so \(L = 100\) kJ/kg.

3. Solid (\(OP\)): \(50\) kJ for \(50^\circ\)C, \(s = \dfrac{50000}{1 \times 50} = 1000\) J/kg K. Liquid (\(QR\)): \(100\) kJ for \(50^\circ\)C, \(s = 2000\) J/kg K. The liquid line is less steep because the liquid needs more heat per degree.

At a steady rate the time is proportional to the heat: \(t \propto mL\). The boiling plateau is longer because \(L_v > L_f\).

\(\dfrac{t_{boil}}{t_{melt}} = \dfrac{L_v}{L_f} = \dfrac{2.26 \times 10^6}{3.36 \times 10^5} \approx 6.7\).

Temperature measures the average kinetic energy of the molecules. During melting the heat goes into pulling the molecules apart against the forces that hold the solid together. It raises their potential energy, not their kinetic energy, so the temperature stays the same.

Heat to warm: \(1 \times 4200 \times 80 = 336000\) J. \(t = \dfrac{336000}{2000} = 168\) s.

Heat to boil away \(0.5\) kg: \(0.5 \times 2.26 \times 10^6 = 1.13 \times 10^6\) J. \(t = \dfrac{1.13 \times 10^6}{2000} = 565\) s.

1. \(C\). For the same rise it needs the most heat, so its line is the steepest. The slope of each line is the heat capacity \(ms\) with \(m = 1\) kg.

2. \(s = \dfrac{Q}{m\Delta T}\) at \(\Delta T = 100\) K: \(A\): \(\dfrac{40000}{100} = 400\); \(B\): \(\dfrac{60000}{100} = 600\); \(C\): \(\dfrac{90000}{100} = 900\) J/kg K.

3. \(Q = 2 \times 400 \times 25 = 20000\) J \(= 20\) kJ.

Heat in one hour \(= 336 \times 3600 = 1.21 \times 10^6\) J. \(m = \dfrac{336 \times 3600}{3.36 \times 10^5} = 3.6\) kg.

\(1\) cal/g \(^\circ\)C \(= \dfrac{4.2 \text{ J}}{10^{-3} \text{ kg K}} = 4200\) J/kg K.

\(80\) cal/g \(= 80 \times 4.2 \times 10^3 = 3.36 \times 10^5\) J/kg. \(540\) cal/g \(= 540 \times 4.2 \times 10^3 = 2.27 \times 10^6\) J/kg.

When hot and cold bodies are mixed, the heat lost by the hot bodies equals the heat gained by the cold bodies. This holds if no heat leaves or enters the system, and if no chemical change or change of state produces or absorbs heat unexpectedly.

\(100(80 - T) = 100(T - 20)\), so \(80 - T = T - 20\) and \(T = 50^\circ\)C. Equal masses of the same liquid meet at the mean temperature.

1. Heat lost \(= 200 \times 1 \times (90 - T)\) cal. Heat gained \(= 100 \times 1 \times (T - 30)\) cal.

2. \(200(90 - T) = 100(T - 30)\), so \(18000 - 200T = 100T - 3000\) and \(300T = 21000\). \(T = 70^\circ\)C.

Heat lost by copper \(= 100 \times 0.1 \times (100 - T) = 10(100 - T)\). Heat gained by water \(= 90(T - 20)\).

\(1000 - 10T = 90T - 1800\), so \(100T = 2800\) and \(T = 28^\circ\)C. The copper counts as only \(10\) g of water, which is why the rise is small.

Heat gained by water and calorimeter \(= (90 + 10) \times 1 \times (40 - 20) = 2000\) cal.

Heat lost by metal \(= 200 \times s \times (140 - 40) = 20000\,s\). So \(20000\,s = 2000\) and \(s = 0.1\) cal/g \(^\circ\)C.

1. Copper conducts heat well, so the vessel and the water reach the same temperature quickly. A thin vessel has a small heat capacity, so it takes little heat itself.

2. The jacket is a poor conductor and the lid stops evaporation and convection. Together they keep heat from leaking to the room, so that heat lost equals heat gained.

3. Stirring mixes the water so that all of it is at one temperature. Otherwise the thermometer reads only the layer around its bulb.

Heat needed to melt all the ice \(= 20 \times 80 = 1600\) cal. Heat available if the water cools to \(0^\circ\)C \(= 80 \times 30 = 2400\) cal. This is more than \(1600\) cal, so all the ice melts.

Then: \(1600 + 20(T - 0) = 80(30 - T)\), so \(1600 + 20T = 2400 - 80T\) and \(100T = 800\). \(T = 8^\circ\)C.

Heat needed to melt all the ice \(= 50 \times 80 = 4000\) cal. Heat available from the water cooling to \(0^\circ\)C \(= 50 \times 40 = 2000\) cal. This is less than \(4000\) cal, so only part of the ice melts and the final temperature is \(0^\circ\)C.

Mass melted \(= \dfrac{2000}{80} = 25\) g. The vessel holds \(25\) g of ice and \(50 + 25 = 75\) g of water, all at \(0^\circ\)C.

Warm the ice to \(0^\circ\)C: \(10 \times 0.5 \times 20 = 100\) cal. Melt it: \(10 \times 80 = 800\) cal. Total \(900\) cal, which the water can supply (\(100 \times 20 = 2000\) cal available).

\(900 + 10T = 100(20 - T)\), so \(110T = 1100\) and \(T = 10^\circ\)C.

Heat gained by water \(= 100 \times (100 - 46) = 5400\) cal.

The final temperature is \(100^\circ\)C, so the condensed steam does not cool at all. Each gram gives only its latent heat, \(540\) cal: \(540m = 5400\) and \(m = 10\) g.

Heat the steam can give while condensing and cooling to \(0^\circ\)C \(= 10 \times 540 + 10 \times 100 = 6400\) cal. Heat needed to melt all the ice \(= 50 \times 80 = 4000\) cal. So all the ice melts and the final temperature is above \(0^\circ\)C.

Heat lost by steam \(= 5400 + 10(100 - T)\). Heat gained by ice \(= 4000 + 50T\).

\(6400 - 10T = 4000 + 50T\), so \(60T = 2400\) and \(T = 40^\circ\)C.

1. Add the four steps: \(10 + 80 + 100 + 540 = 730\) cal.

2. Down two steps: \(540 + 100 = 640\) cal.

3. Two steps up for each gram: \(10 + 80 = 90\) cal. For \(5\) g: \(450\) cal.

Heat lost by \(A\) \(= 100 \times 0.5 \times (60 - T) = 50(60 - T)\). Heat gained by \(B\) \(= 200 \times 0.25 \times (T - 20) = 50(T - 20)\).

The heat capacities are equal, \(50\) cal/\(^\circ\)C each, so \(T\) is the mean: \(T = 40^\circ\)C.

Let each mass be \(m\). The hot water loses \(m(60 - T)\). The cold waters gain \(m(T - 10) + m(T - 20)\).

\(60 - T = 2T - 30\), so \(3T = 90\) and \(T = 30^\circ\)C, the mean of the three.

1. The solid starts at \(100^\circ\)C, the water at \(20^\circ\)C. Both curves meet at \(40^\circ\)C.

2. Equal masses \(m\): \(m\,s\,(100 - 40) = m \times 1 \times (40 - 20)\), so \(60s = 20\) and \(s = \dfrac{1}{3}\) cal/g \(^\circ\)C.

3. The same heat flows out of the solid and into the water each second. Water has the larger specific heat, so the same heat changes its temperature less.

Melting the ice needs \(30 \times 80 = 2400\) cal; the drink can supply \(300 \times 30 = 9000\) cal, so all the ice melts.

\(2400 + 30T = 300(30 - T)\), so \(330T = 6600\) and \(T = 20^\circ\)C.

The calorimeter and its water count as \(80 + 20 = 100\) g of water at \(20^\circ\)C.

\(100(70 - T) = 100(T - 20)\), so \(T = 45^\circ\)C.

The aluminium beaker. For the same mass and the same fall in temperature, the heat given out is \(ms\Delta T\). Aluminium has twice the \(s\) of copper. The aluminium ball gives out about twice as much heat, so its water warms more.

With equal masses, \(m s_1 (60 - 40) = m s_2 (40 - 20)\), so \(s_1 = s_2\). The final temperature is the mean only when the heat capacities are equal.

Heat needed to melt all the ice \(= 100 \times 80 = 8000\) cal. Heat available from the water cooling to \(0^\circ\)C \(= 100 \times 80 = 8000\) cal. The two are exactly equal.

So all the ice just melts and the final temperature is \(0^\circ\)C. The vessel holds \(200\) g of water at \(0^\circ\)C.

Each gram of steam gives \(540\) cal on condensing and \(100\) cal on cooling to \(0^\circ\)C: \(640\) cal. So \(2\) g gives \(1280\) cal.

Ice melted \(= \dfrac{1280}{80} = 16\) g. The vessel holds \(100 - 16 = 84\) g of ice and \(100 + 16 + 2 = 118\) g of water, all at \(0^\circ\)C.

1. The curve passes through \(T = 20^\circ\)C at \(m = 20\) g. Check: \(80 \times 20 + 20 \times 20 = 2000\) cal gained, and \(100 \times (40 - 20) = 2000\) cal lost.

2. The curve reaches \(0^\circ\)C at \(m = 50\) g. Check: the water can give \(100 \times 40 = 4000\) cal, which melts \(\dfrac{4000}{80} = 50\) g.

3. Only \(50\) g melts. The vessel holds \(10\) g of ice and \(150\) g of water, all at \(0^\circ\)C.

Each gram of steam gives \(540 + 100 = 640\) cal in becoming water at \(0^\circ\)C. Melting \(100\) g of ice needs \(8000\) cal.

\(m = \dfrac{8000}{640} = 12.5\) g.