Questions: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18

1. In a physics experiment, 425 g of lead shot at 220.0 °C is added to 550 g of water at 16.0 °C in an insulated thermos. What is the final temperature of the lead and water?

   The heat lost by the lead must be absorbed by the water. The insulated thermos prevents losses to the surroundings, so the heat energy is conserved. Thus

   Q_Pb + Q_W = 0.

   As long as there is no change of state, Q = mcΔT = mc(T_f - T_i). The temperatures should be in Kelvin, so 220.0 °C = 220.0 K + 273.16 K = 493.16 K and 16.0 °C = 16.0 K + 273.16 K = 289.16 K. The lead and water will have the same temperature T_f when they reach equilibrium.

   Thus our equation becomes

   m_Pbc_Pb(T_f - T_{i Pb}) + m_Wc_W(T_f - T_{i W}) = 0.

   Getting the terms in T_f isolated on the left hand side yields,

   (m_Pbc_Pb + m_Wc_W)T_f = m_Pbc_PbT_{i Pb} + m_Wc_WT_{i W} .

   Using the given values

   

   So the lead and water reach equilibrium at 20.7 °C. Notice that in doing this problem, we took the precaution of converting all temperatures from Celsius to Kelvin.

   

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2. In another physics experiment, 75 g of ice at -20.0 °C is added to 1350 g of water at 80.0 °C in an insulated thermos. What is the final temperature of the water?

   The heat lost by the ice must be absorbed by the water. The insulated thermos prevents losses to the surroundings, so the heat energy is conserved. Thus

   Q_ice + Q_W = 0.

   The ice must warm from -20 °C (253.16 K) up to 0 °C (273.16 K), then melt, and then the ice water must warm to the final temperature. At the same time the other water cools from 80 °C (353.16 K). The ice water and the other water will have the same temperature T_f when they reach equilibrium. When a material is in one state, Q = mcΔT = mc(T_f - T_i). When ice melts the heat required is given by Q_melting = mL_F.

   In this case, therefore, we have

   Q_{ice warming} + Q_{ice melting} + Q_{icewater warming} + Q_W = 0,

   or

   m_icec_ice(273.16 K - T_{ice 0}) + m_iceL_F + m_icec_W(T_f - 273.16 K) + m_Wc_W(T_f - T_{W 0}) = 0 .

   We isolate T_f

   (m_ice + m_W)c_WT_f = m_icec_W(273.16 K) + m_Wc_WT_{W 0} - m_icec_ice(273.16 K - T_{ice 0}) - m_iceL_F

   Using the given values

   

   This reduces to

   (5970.75 J/K)T_f = (45481.14 + 1997649.54 - 3330 - 24975) J .

   So

   T_f = (2014825.68 J)/(5970.75 J/K) = 337.45 K = 64.29 °C .

   The system reaches equilibrium at 64.3 °C.

   

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3. A student's 250 g coffee has cooled to 30 °C while he has been studying. He decides to reheat the coffee by placing it in his 300 W microwave. How long does he need to set the microwave for, if he wishes to heat the coffee to 70 °C. Treat the coffee as being water and assume it absorbs all the microwave's energy. Recall that power is energy output over time.

   We are told to assume

   E_microwave = E_water .

   Since the rate at which energy is emitted by the microwave is P = 300 W,

   E_microwave = (300 W)t ,

   where t is the elapsed time. The energy that the water absorbs is given by

   E = Q = mc_waterΔT _.

   Combining the results, yields

   (300 W)t = mc_waterΔT _.

   We then find the required time to be

   t = mc_wΔT / 300 W = (0.250 kg)(4190 J/kg-K)(70 °C - 30 °C)/(300 W) = 139.7 s_.

   It will take the coffee two minutes and twenty seconds to warm up sufficiently.

   

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4. When you exercise, you produce lots of thermal energy internally. If you don't dissipate the heat, your core temperature would rise dangerously high. During heavy exercise you may need to dissipate hundreds of extra watts of thermal energy. Suppose that your core is at 37 °C and your skin is at 34 °C. Assume that you have on average 3 cm of fat and a surface area of 1.5 m². Take the thermal conductivity of body fat to be 0.20 J/(s·m·C). What is the rate of heat loss through your skin? Will you need other ways of cooling down?

   The heat loss through a conductor is given by Q = κA(T_in - T_out)/L, where A is the surface area and L is the thickness of the conductor. We are given κ so the result is

   Q = [0.20 J/(s·m·C)](1.5 m²)(37 °C − 34 °C)/(0.03 m) = 30 J.

   Not a very large number and not a good way to cool.

   

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5. Another way your body has of cooling is by flushing - warm blood is sent to the surface arteries and this blood cools down quicker because there is so little fat in the way. The cool blood returns to the core to collect more thermal energy and to repeat the process. An even more powerful way of cooling down is sweating. The sweat (basically water at skin temperature) will evaporate carrying energy away from the body. If a person is exercising and producing an extra 250 W of thermal energy, what mass of sweat does the person produce in an hour? Note that evaporation occurs best into dry air, so when it is very humid sweating does not cool very effectively.

   Evaporation carries away Q = mL_v where L_v = 22.6 × 10⁵ J/kg. Let's assume all the sweat evaporates. The thermal energy it must carry away in one hour is Q = (250 W) × 3600 s = 9 × 10⁵ J. The mass of evaporated sweat is thus

   m = Q/L_v = (9 × 10⁵ J)/(9 × 10⁵ J/kg) = 0.40 kg or 0.40 L.

   

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6. A wall has a area of 10 m². It is 3.00 cm thick polyurethane (κ = 0.024 J/s·m·K) on 1.00 cm of wood (κ = 0.010 J/s·m·K). The interior is 22 °C and the exterior is 5 °C. What is the temperature at the boundary between the polyurethane and the wood?

   The heat flow through the polyurethane must equal the heat flow through the wood. Let's call the temperature at the boundary T_boundary.

   κ_polyurethaneA(T_in - T_boundary)/L_polyurethane = κ_woodA(T_boundary - T_out)/L_wood

   A is common and can be eliminated.

   κ_polyurethane(T_in - T_boundary)/L_polyurethane = κ_wood(T_boundary - T_out)/L_wood

   Next collect terms with T_boundary on the same side.

   κ_polyurethaneT_in/L_polyurethane + κ_woodT_out/L_wood = κ_woodT_boundary/L_wood + κ_polyurethaneT_boundary/L_polyurethane

   Using the values for κ for polyurethane and wood given in the text and the other values given in the problem we find T_boundary

   (0.024 J/(s·m·C))(22 °C)/(0.03 m) + (0.1 J/(s·m·C))(5 °C)/(0.01 m) = [(0.1 J/(s·m·C))/(0.01 m) + (0.024 J/(s·m·C))/(0.03 m)]T_boundary

   This reduces to

   (17.6 J/(s·m²) + 50 J/(s·m²)) = [10 J/(s·m²·C) + 0.8 J/(s·m²·C)]T_boundary

   This yields T_boundary = 6.26 °C.

   

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7. Triple-glazed glass consists of three 0.25 cm thick panes of glass separated by two air gaps of 1.00 cm thickness. What is the effective R-value of this window? Note κ_glass = 1.0 W/m·K and κ_air = 0.026 W/m·K.

   Examining the appropriate table from the text, the thermal conductivity of glass is K_glass = 1.0 W/m-K and K_air = 0.026 W/m-K.

   The effective R-value is given by the formula

   R = L₁/K₁ + L₂/K₂ + L₃/K₃ + ... ,

   where T is the thickness of the layer and K is its thermal conductivity. Thus

   R = 3L_glass/K_glass + 2L_air/K_air = (30.025)/(1) + (20.010)/(0.026) = 0.844 m²K/W .

   The wind has an effective R-value of 0.84 m²K/W .
   

   

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8. A cabin has a surface area of 90 m². The cabin is made of 2.5 cm thick wood with a thermal conductivity of 0.13 W/m-K. What is the R-value of the wood? If the exterior temperature is -10 °C, what would have to be the power of a space heater to keep the interior at 17 °C? If the cabin was also insulated with 2.4 cm of fibreglass (k = 0.048 W/m-K), what would be the necessary power of the space heater? What would be the temperature of the interface between the fibreglass and the wood?

   The effective R-value is given by the formula

   R = L₁/K₁ + L₂/K₂ + L₃/K₃ + ... ,

   so

   R = (0.025 m)/(0.13 W/m-K) = 0.1923 m²K/W .

   The heat loss through a conductor is given by H = A(T_in - T_out)/R, where A is the surface area. Thus

   H = (90 m²)[17 °C - (-10 °C)]/( 0.1923 m²K/W) = 1.26 × 10⁴ W .

   We didn't need to convert the temperature from Celsius to Kelvin because we are dealing with a difference in temperature. Thus the heater would have to use power at the rate 1.26 × 10⁴ W to keep the cabin at 17 °C.

   With the insulation the new R-value is

   R = (0.025 m)/(0.13 W/m-K) + (0.024 m)/( 0.048 W/m-K) = 0.7131 m²K/W .

   Thus

   H_insulation = (90 m²)[17 °C - (-10 °C)]/( 0.7131 m²K/W) = 3.408 × 10³ W .

   Hence the heater need only supply 3400 W of power.

   At steady state the temperature at the interface between the insulation and the wood will clearly be some temperature, T_i, between 17 C and -10 °C. As well energy must be conserved, the amount of heat energy passing through the insulation must pass through the wood,

   H_insulation = H_wood .

   Using our formula

   A(T_in - T_i)/(L_insulation/K_insulation) = A(T_i - T_out)/(L_wood/K_wood) .

   We eliminate the common factor A, and collect terms involving T_i to get

   [(K_wood/L_wood) + (K_insulation/L_insulation)]T_i = -[(K_wood/L_wood)T_out - (K_insulation/L_insulation)T_in] .

   or

   

   This reduces to

   T_i = (52 + 34) / (5.2 + 2) = 11.94 °C .

   The interface is at 11.9 °C.

   

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9. A matte (i.e. dull) black cube has a side length of 20.0 cm. It is suspended in 20 °C still air. It is absorbing sunlight at a rate of 400 W/m² from the Sun directly overhead. What will be its final equilibrium temperature if it only loses energy by radiation? If the cube was shiny white, and reflected 60% of the sunlight, what would its final temperature be?

   Since still air is a poor conductor we need only worry about radiation. The top side (area = L²) only of the cube absorbs energy from the sun without reflection. All six sides (area = 6L²) receive energy form the surrounding environment which is at 20 °C. All six sides emit energy at the currently unknown temperature of the box. The energy absorbed must equal the energy readmitted and thus our equation is

   I_sunL² + σ(6L²)T_air⁴ = σ(6L²)T_cube⁴

   Noice that L² cancels out; the dimensions of the cube do not matter. Also temperature must be in Kelvin not Celsius. Our equation becomes

   I_sun + 6σT_air⁴ = 6σT_cube⁴

   Or

   I_sun/6σ + T_air⁴ = T_cube⁴

   Substituting the values given

   (400 W/m²)/6(5.6703×10^-8 W/m²K⁴) + (20 + 273.15 K)⁴ = T_cube⁴

   We find T_cube = 304.18 K or 31.0 °C.

   The white cube absorbs only 40% of the sunlight so, we need only slightly modify our equation to

   0.4I_sun/6σ + T_air⁴ = T_cube⁴

   Substituting the values given

   (400 W/m²)/6(5.6703×10^-8 W/m²K⁴)(0.5) + (20 + 273.15 K)⁴ = T_cube⁴

   We find T_cube = 297.71 K or 24.6 °C.

   

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10. If you stand naked in a room, your skin and the walls of the room will exchange heat by radiation. Suppose the temperature of your skin is 33 °C, the surface area of your skin is 1.5 m², and the temperature of the walls is 15 °C. Assume that your body and the walls act as blackbodies.
    (a) What is the rate at which your body radiates heat?
    (b) What is the rate at which your skin absorbs heat?
    (c) What is the net rate of your loss of heat?
    (d) How many Oh Henry! bars (1338 kJ per bar) would you have to eat in a day to survive?
    

    (a) The rate at which a body radiates energy because of its temperature is given by the Stephan-Boltzmann Law,
    P = σεAT⁴, where T is the temperature of the body and A is its area. For a blackbody, the ε = 1. In this problem

    P_emitted = (5.6703×10^-8 W/m²K⁴)(1)(1.5 m²)(273.16 K + 33 K) = 745.7 W .

    (b) The rate at which a body absorbs energy because of the temperature of its surroundings is also given by the Stephan-Boltzmann Law, P = σεAT⁴, where T is the temperature of the surroundings and A is still the area of the body. For a blackbody, the ε = 1. In this problem

    P_absorbed = (5.6703×10^-8 W/m²K⁴)(1)(1.5 m²)(273.16 K + 15 K) = 585.2 W .

    (c) The net loss of power is thus, ΔP = P_e - P_a = 160.5 W.

    In one day the energy loss would be

    E = ΔP × t = (160.5 W)(24 h/d 3600 s/h) = 1.387 × 10⁸ J .

    (d) Since we are given the energy in the candy bar, the number of bars required is

    n = E / E_bar = (1.387 × 10⁸ J)/( 1.387 × 10⁸ J) = 10.4 .

    So you would need to eat 10.4 bars just to stay warm.

    Of course, people aren't blackbodies, they wear insulating clothes and have an insulating layer of fat and skin. So ε < 1, and the energy requirement above is an overestimate. We can estimate ε by noting that a sedentary person needs about 1600 kilocalories per day, or 6700 KJ or 5 Oh!Henry bars. Thus ε must be around 0.5 .

    

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11. Assume the naked person in the above question would feel comfortable at a metabolic rate of 100 W. What would the room temperature have to be for him to feel comfortable?

    The person is losing thermal energy radiatively as given by the formula

    P_net = σεAT_skin⁴ − σεAT_environment⁴

    Now ε = 1 for a blackbody. We know A = 1.5 m² and σ = 5.6703×10^-8. The skin temperature is T_skin = 34 + 273 K = 307 K. For the person to be comfortable P_net = 100 W.

    We can rearrange the above equation for the comfortable ambient temperature

    T_environment = [T_skin⁴ − P_net/σεA ]^1/4 = 296 K = 23 C

    

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12. For the same person in Question 10, they decide to put on a wool catsuit or leotard (i.e. something tight and clingy so there is no layer of air), κ = 0.040 W/m·K, to stay comfortable. What will be the temperature of the surface of the catsuit? How thick must the catsuit be? Note that the net thermal energy from the person must pass through the catsuit by conduction before being radiated away at the surface of the catsuit.

    The thermal energy, 100 W, conducts through cloth and obeys the formula

    P_net = κA[T_skin − T_{catsuit exterior}]/L

    but we don't know the exterior temperature of the catsuit but it will be below the person's skin temperature of 34 C.

    By conservation of energy, the same power is radiated from the catsuit according to the formula

    P_net = σεAT_{catsuit exterior}⁴ − σεAT_environment⁴

    From this second formula we can find the catsuit exterior temperature (noting that that T_environment = 15 + 273 K = 288 K, A = 1.5 m², σ = 5.6703×10^-8, and for blackbodies ε = 1)

    T_{catsuit exterior} = [T_environment⁴ + P_net/εσA ]^1/4 = 299.6 K = 26.6 C

    With this temperature we can return to the first formula and solve for L using κ = 0.040 W/m·K.

    L = κA[T_skin − T_{catsuit exterior}]/P_net = (0.040)(1.5)(34 - 26.6)/100 = 0.0044 m.

    A catsuit of 4.4 mm thickness will do the job.

    

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13. Suppose the room temperature is Question 10 is 25 C, slightly warmer than comfortable. The person flushes to cool down. In flushing, warm blood is brought to the surface of the skin raising skin temperature. The blood returning from the skin to the core will be cooler than before flushing will also cool the person but ignore that effect. Find the new skin temperature if the person is comfortable (has a metabolic rate of 100 W)?

    Assuming all radiate cooling

    P_net = σεAT_{flushed skin}⁴ − σεAT_environment⁴

    Solving for the temperature of the flushed skin and noting T_environment = 25 + 273 K = 298 K, A = 1.5 m², σ = 5.6703×10^-8, and for blackbodies ε = 1, we find

    T_{flushed skin} = [T_environment⁴ + P_net/εσA ]^1/4 = 308.5 K = 35.5 C

    Since there is the extra cooling of cool blood returning to the core, the flushed skin temperature would actually be lower.

    

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14. Suppose the room temperature is Question 10 is 30 C, warmer than comfortable. The person sweats to cool off. How many grams of sweat will be produced in one hour if the person's metabolic rate is to stay at 100 W?

    At this higher temperature the radiative cooling is only

    P_net = σεAT_{flushed skin}⁴ − σεAT_environment⁴ = 38.6 W

    where I have used T_skin = 34 + 273 K = 307 K, T_environment = 30 + 273 K = 303 K, A = 1.5 m², σ = 5.6703×10^-8, and for blackbodies ε = 1.

    So an extra 100 W - 38.6 W = 61.4 W must be lost by sweating

    Sweating involved latent heat Q = mL_F and P = Q/t, so to find the mass of sweat in one hour, P = mL_F/t can be rearraged to give

    m = Pt/L_F = (61.4 W)(3600 s)/(2260000 J/kg) = 0.098 kg

    So the person sweats 664 g. Note we are ignoring the fact that the person would flush thus raising the skin temperature and threby reducing the amount of sweat produced.

    

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15. The sun is 150 × 10⁹ m from the earth. The surface temperature of the sun is 5776 K. The flux (or power per unit area) of sunlight at the earth is 1.34 × 10³ W/m².
    (a) Determine the total power emitted by the sun. The surface area of a sphere is 4πr².
    (b) Calculate the radius of the sun. Assume that the sun is a blackbody.
    (Note - Astronomers determine the surface temperature of a star from its colour. In this way they can determine the radii of stars by measuring the flux of starlight.)
    

    (a) The Sun is a sphere and it radiates energy uniformly in all directions. If we consider a sphere of radius R, where R is the distance from the Sun to the Earth, each square metre of that enormous sphere must intercept the same energy as on earth. Hence the power emitted by the Sun must be
    

    P_emitted = 4πR² × (1.34 × 10³ W/m²) = 4π(150 ×10⁹)(1.34 × 10³) = 3.7888 × 10²⁶ W .

    So the Sun emits energy at the rate of 3.79 × 10²⁶ W.

    (b) According to the Stephan-Boltzmann Law, the power emitted by the Sun is also

    P_emitted = σεAT⁴ .

    The surface area of the Sun is A = 4π(R_S)². Hence

    P_emitted = 4π(R_S)²T⁴ .

    We treat the Sun as a blackbody so ε = 1. Thus we can find the radius of the Sun,

    R_Sun = [P_e/4pσεT⁴]^½ = 6.91 × 10⁸ m.

    So the radius of the Sun is 200 times smaller than the distance from the Sun to the Earth.

    

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16. Suppose that your core is at 37 °C and the ambient or air temperature is 5 °C. The temperature of your skin is determined by how the rate energy is conducted through your skin from the core to your skin and the rate at which you can emit thermal radiation to the environment. (We are neglecting the insulating properties of the air and any wind chill). Assume that you have on average 3 cm of fat and a surface area of 1.5 m². Take the thermal conductivity of body fat to be 0.20 J/(s·m·°C). Find your skin temperature to 3 significant figures. You will need to try some values (say 25 °C and 20 °C) and then try new values to get to the correct answer. A spreadsheet let's you do this easily. Assume you are a blackbody.

    The rate at which energy is conducted out of the core is

    P = A(T_interior − T_skin) / R where R = L/κ = .03 m / 0.20 J/(s·m·°C) = 0.25 (s·m²·°C) /J

    The rate that energy is emitted from your skin is

    P = σA(T_skin⁴ − T_env⁴). Remember these temperatures must be in Kelvin.

    Let's calculate these powers at different temperatures.

    Tskin (°C)	Tskin (Kelvin)	Pconduction (W)	Pradiation (W)
    30	303.16	70	209.2452
    25	298.16	120	163.0088
    20	293.16	170	119.0409
    15	288.16	220	77.26599

    The skin temperature is set when the two powers equal. From this table we see that this temperature must be between 25 and 20 degrees. Continuing

    Tskin (°C)	Tskin (Kelvin)	Pconduction (W)	Pradiation (W)
    25	298.16	120	163.0088
    24	297.16	130	154.0362
    23	296.16	140	145.1537
    22	295.16	150	136.3607
    21	294.16	160	127.6566
    20	293.16	170	119.0409

    Now it appears that the answer is between 23 and 22 degrees. So continue again

    Tskin (°C)	Tskin (Kelvin)	Pconduction (W)	Pradiation (W)
    22.8	295.96	142	143.3879
    22.6	295.76	144	141.6258
    22.4	295.56	146	139.8672
    22.2	295.36	148	138.1121
    22.72	295.88	142.8	142.7

    Final skin temperature is about 22.7 °C.

    

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17. The person in the previous problem will cool rapidly since a typical 70 kg person generates 100 W of heat. To keep warm, one can engage in physical activity. How much heat must he generate? What amount of physical activity (work) must he do? Luckily this person is hill climbing, at what rate (in metres per second and metres per hour) must he climb? Assume that he is 25% efficient?

    He needs to generate 142.8 W − 100 W = 42.8 W of extra heat.

    Since efficiency is given by e = (Work/time / Heat/time), then Work/time = 0.25 × 42.8 W = 10.7 W.

    A climbing person is gaining gravitation energy, mgh. So W/t = mgh/t and we are looking for h/t.

    Our equation is h/t = (W/t) / mg = 10.7 / (70)(9.81) = 0.0155 m/s or 56 m/h which is not very strenuous.

    

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18. A kg of ice fills a cube 10 cm on each side. Suppose the ice is at 0 C and the ambient temperature is 20 C. How long will it take to melt completely to water at 0 C? Treat the ice as having ε = 0.95.

    The latent heat Q = mL_F must be absorbed from the surrounding. The rate that this happens Q/t = mL_F/t must equated the net absorbed radiative power.

    mL_F/t = σεAT_environment⁴ − σεAT_ice⁴

    We know m = 1.0 kg, area A = 6 × (0.1 m) × (0.1 m) = 0.06 m², L_F = 333 kJ/kg, T_ice = 0 + 273 K = 273 K, T_environment = 20 + 273 K = 293 K, and σ = 5.6703×10^-8. We are given ε = 0.95. So rearranging out equation to find t,

    t = mL_F / σεA[T_environment⁴ − T_ice⁴] = (1)(333000) / (0.95)(5.6703×10^-8)(0.06)[(293)⁴ − (273)⁴] = 56750 s.

    It takes almost 16 hours to melt!

    

Questions? mike.coombes@kpu.ca