Temperature and heat(theory questions) Flashcards
What does it mean to say that two systems are in thermal equilibrium?
They are at the same temperature, and if they are placed in contact, no net heat flows between them.
If a thermometer is allowed to come to equilibrium with the air, and a glass of water is not in equilibrium with the air, what will happen to the thermometer reading when it is placed in the water?
The reading will change.
Pouring cold water into hot glass or ceramic cookware can easily break it. What causes the breaking? Explain why Pyrex, a glass with a small coefficient of linear expansion, is less susceptible.
The cold water cools part of the inner surface, making it contract, while the rest remains expanded. The strain is too great for the strength of the material. Pyrex contracts less, so it experiences less strain.
Does it really help to run hot water over a tight metal lid on a glass jar before trying to open it? Explain your answer.
In principle, the lid expands more than the jar because metals have higher coefficients of expansion than glass. That should make unscrewing the lid easier. (In practice, getting the lid and jar wet may make gripping them more difficult.)
Calculate the length of a 1-meter rod of a material with thermal expansion coefficient alpha when the temperature is raised from 300 K to 600 K. Taking your answer as the new initial length, find the length after the rod is cooled back down to 300 K. Is your answer 1 meter? Should it be? How can you account for the result you got?
After being heated, the length is (1 + 300 alpha) (1 m). After being cooled, the length is (1 - 300 alpha) (1 + 300 alpha) (1 m). That answer is not 1 m, but it should be. The explanation is that even if alpha is exactly constant, the relation delta_L = alpha L delta_T is strictly true only in the limit of small delta_T. Since alpha values are small, the discrepancy is unimportant in practice.
How is heat transfer related to temperature?
Temperature differences cause heat transfer.
When heat transfers into a system, is the energy stored as heat? Explain briefly.
No, it is stored as thermal energy. A thermodynamic system does not have a well-defined quantity of heat.
A pressure cooker contains water and steam in equilibrium at a pressure greater than atmospheric pressure. How does this greater pressure increase cooking speed?
It raises the boiling point, so the water, which the food gains heat from, is at a higher temperature.
Can carbon dioxide be liquefied at room temperature (20°C)? If so, how? If not, why not? (See the phase diagram in the preceding problem.)
Yes, by raising the pressure above 56 atm.
Heat transfer can cause temperature and phase changes. What else can cause these changes?
Work
What is the temperature of ice right after it is formed by freezing water?
0°C (at or near atmospheric pressure)
What effect does condensation on a glass of ice water have on the rate at which the ice melts? Will the condensation speed up the melting process or slow it down?
Condensation releases heat, so it speeds up the melting.
In winter, it is often warmer in San Francisco than in Sacramento, 150 km inland. In summer, it is nearly always hotter in Sacramento. Explain how the bodies of water surrounding San Francisco moderate its extreme temperatures.
Because of water’s high specific heat, it changes temperature less than land. Also, evaporation reduces temperature rises. The air tends to stay close to equilibrium with the water, so its temperature does not change much where there’s a lot of water around, as in San Francisco but not Sacramento.
In a physics classroom demonstration, an instructor inflates a balloon by mouth and then cools it in liquid nitrogen. When cold, the shrunken balloon has a small amount of light blue liquid in it, as well as some snow-like crystals. As it warms up, the liquid boils, and part of the crystals sublime, with some crystals lingering for a while and then producing a liquid. Identify the blue liquid and the two solids in the cold balloon. Justify your identifications using data from the table (Melting Point (°C), Boiling Point (°C) for: Nitrogen -210.0, -195.8; Oxygen -218.8, -183.0; Water 0.00, 100.0).
The liquid is oxygen, whose boiling point (-183.0°C) is above that of nitrogen (-195.8°C) but whose melting point (-218.8°C) is below the boiling point of liquid nitrogen. The crystals that sublime are carbon dioxide, which has no liquid phase at atmospheric pressure. The crystals that melt are water, whose melting point (0.00°C) is above carbon dioxide’s sublimation point. The water came from the instructor’s breath.
Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while the surface only a few centimeters away is safe to touch. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?
It spreads the heat over the area above the heating elements, evening the temperature there, but does not spread the heat much beyond the heating elements.
When our bodies get too warm, they respond by sweating and increasing blood circulation to the surface to transfer thermal energy away from the core. What effect will those processes have on a person in a 40.0°C hot tub?
Increasing circulation to the surface will warm the person, as the temperature of the water is warmer than human body temperature. Sweating will cause no evaporative cooling under water or in the humid air immediately above the tub.
One way to make a fireplace more energy-efficient is to have room air circulate around the outside of the fire box and back into the room. Detail the methods of heat transfer involved.
Heat is conducted from the fire through the fire box to the circulating air and then convected by the air into the room (forced convection).
When watching a circus during the day in a large, dark-colored tent, you sense significant heat transfer from the tent. Explain why this occurs.
The tent is heated by the Sun and transfers heat to you by all three processes, especially radiation.
Your house will be empty for a while in cold weather, and you want to save energy and money. Should you turn the thermostat down to the lowest level that will protect the house from damage such as freezing pipes, or leave it at the normal temperature? (If you don’t like coming back to a cold house, imagine that a timer controls the heating system so the house will be warm when you get back.) Explain your answer.
Turn the thermostat down. To have the house at the normal temperature, the heating system must replace all the heat that was lost. For all three mechanisms of heat transfer, the greater the temperature difference between inside and outside, the more heat is lost and must be replaced. So the house should be at the lowest temperature that does not allow freezing damage.
Why are thermometers that are used in weather stations shielded from the sunshine? What does a thermometer measure if it is shielded from the sunshine? What does it measure if it is not?
If shielded, it measures the air temperature. If not, it measures the combined effect of air temperature and net radiative heat gain from the Sun.
Broiling is a method of cooking by radiation, which produces somewhat different results from cooking by conduction or convection. A gas flame or electric heating element produces a very high temperature close to the food and above it. Why is radiation the dominant heat-transfer method in this situation?
Air is a good insulator, so there is little conduction, and the heated air rises, so there is little convection downward.
The height of the Washington Monument is measured to be 170.00 m on a day when the temperature is 35.0°C. What will its height be on a day when the temperature falls to -10.0°C? Although the monument is made of limestone, assume that its coefficient of thermal expansion is the same as that of marble (alpha = 2.5 x 10^-6 /°C). Give your answer to five significant figures.
L = L_0 + delta_L = L_0 (1 + alpha delta_T) = 170 m [1 + (2.5 x 10^-6 /°C) (-45.0°C)] = 169.98 m. (Answer rounded to five significant figures to show the slight difference in height.)
What is the change in length of a 3.00-cm-long column of mercury if its temperature changes from 37.0°C to 40.0°C, assuming the mercury is constrained to a cylinder but unconstrained in length? Your answer will show why thermometers contain bulbs at the bottom instead of simple columns of liquid.
We use beta instead of alpha since this is a volume expansion with constant surface area. Therefore: delta_L = alpha L delta_T = (6.0 x 10^-5 /°C) (0.0300 m) (3.00°C) = 5.4 x 10^-6 m.
(a) Suppose a meter stick made of steel and one made of aluminum are the same length at 0°C. What is their difference in length at 22.0°C? (b) Repeat the calculation for two 30.0-m-long surveyor’s tapes. (alpha_al = 2.5 x 10^-5 /°C, alpha_steel = 1.2 x 10^-5 /°C)
a. delta_L_Al - delta_L_steel = (alpha_Al - alpha_steel) L_0 delta_T = (2.5 x 10^-5 /°C - 1.2 x 10^-5 /°C) (1.00 m) (22°C) = 2.9 x 10^-4 m; b. By the same method with L_0 = 30.0 m, we have delta_L = 8.6 x 10^-3 m.
Most cars have a coolant reservoir to catch radiator fluid that may overflow when the engine is hot. A radiator is made of copper and is filled to its 16.0-L capacity when at 10.0°C. What volume of radiator fluid will overflow when the radiator and fluid reach a temperature of 95.0°C, given that the fluid’s volume coefficient of expansion is beta = 400 x 10^-6 /°C? (alpha_copper = 17 x 10^-6 /°C) (Your answer will be a conservative estimate, as most car radiators have operating temperatures greater than 95.0°C).
delta_V = 0.475 L
The density of water at 0°C is very nearly 1000 kg/m^3 (it is actually 999.84 kg/m^3), whereas the density of ice at 0°C is 917 kg/m^3. Calculate the pressure necessary to keep ice from expanding when it freezes, neglecting the effect such a large pressure would have on the freezing temperature. (This problem gives you only an indication of how large the forces associated with freezing water might be.)
If we start with the freezing of water, then it would expand to (1 m^3) (1000 kg/m^3 / 917 kg/m^3) = 1.09 m^3 = 1.98 x 10^8 N/m^2 of ice.
The same heat transfer into identical masses of different substances produces different temperature changes. Calculate the final temperature when 1.00 kcal of heat transfers into 1.00 kg of the following, originally at 20.0°C: (a) water; (b) concrete; (c) steel; (d) mercury.
Q = m c delta_T => delta_T = Q / m c, a. 21.0°C; b. 25.0°C; c. 29.3°C; d. 50.0°C
On a hot day, the temperature of an 80,000-L swimming pool increases by 1.50°C. What is the net heat transfer during this heating? Ignore any complications, such as loss of water by evaporation.
5.02 x 10^8 J
A 0.250-kg block of a pure material is heated from 20.0°C to 65.0°C by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed.
Q = m c delta_T => c = Q / m delta_T = 1.04 kcal / (0.250 kg) (45.0°C) = 0.0924 kcal/kg·°C. It is copper.
(a) The number of kilocalories in food is determined by calorimetry techniques in which the food is burned and the amount of heat transfer is measured. How many kilocalories per gram are there in a 5.00 g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a 54.9°C temperature increase? Assume the process takes place in an ideal calorimeter, in other words a perfectly insulated container. (b) Compare your answer to the following labeling information found on a package of dry roasted peanuts: a serving of 33 g contains 200 calories. Comment on whether the values are consistent.
a. Q = m_w c_w delta_T + m_Al c_Al delta_T = (m_w c_w + m_Al c_Al) delta_T; Q = [(0.500 kg) (1.00 kcal/kg·°C) + (0.100 kg) (0.215 kcal/kg·°C)] (54.9°C) = 28.63 kcal; Q / m_p = 28.63 kcal / 5.00 g = 5.73 kcal/g; b. Q / m_p = 200 kcal / 33 g = 6 kcal/g, which is consistent with our results to part (a), to one significant figure.
In a study of healthy young men, doing 20 push-ups in 1 minute burned an amount of energy per kg that for a 70.0-kg man corresponds to 8.06 calories (kcal). How much would a 70.0-kg man’s temperature rise if he did not lose any heat during that time?
0.139°C
Repeat the preceding problem, assuming the water is in a glass beaker with a mass of 0.200 kg, which in turn is in a calorimeter. The beaker is initially at the same temperature as the water. Before doing the problem, should the answer be higher or lower than the preceding answer? Comparing the mass and specific heat of the beaker to those of the water, do you think the beaker will make much difference?
It should be lower. The beaker will not make much difference: 16.3°C
A bag containing 0°C ice is much more effective in absorbing energy than one containing the same amount of 0°C water. (a) How much heat transfer is necessary to raise the temperature of 0.800 kg of water from 0°C to 30.0°C? (b) How much heat transfer is required to first melt 0.800 kg of 0°C ice and then raise its temperature? (c) Explain how your answer supports the contention that the ice is more effective.
a. 1.00 x 10^5 J; b. 3.68 x 10^5 J; c. The ice is much more effective in absorbing heat because it first must be melted, which requires a lot of energy, and then it gains the same amount of heat as the bag that started with water. The first 2.67 x 10^5 J of heat is used to melt the ice, then it absorbs the 1.00 x 10^5 J of heat as water.
Condensation on a glass of ice water causes the ice to melt faster than it would otherwise. If 8.00 g of vapor condense on a glass containing both water and 200 g of ice, how many grams of the ice will melt as a result? Assume no other heat transfer occurs. Use L_v for water at 37°C as a better approximation than L_v for water at 100°C.
58.1 g
On a certain dry sunny day, a swimming pool’s temperature would rise by 1.50°C if not for evaporation. What fraction of the water must evaporate to carry away precisely enough energy to keep the temperature constant?
Let M be the mass of pool water and m be the mass of pool water that evaporates. M c delta_T = m L_v(37°C) => m / M = c delta_T / L_v(37°C) = (1.00 kcal/kg·°C) (1.50°C) / 580 kcal/kg = 2.59 x 10^-3; (Note that L_v for water at 37°C is used here as a better approximation than L_v for 100°C water.)
In 1986, an enormous iceberg broke away from the Ross Ice Shelf in Antarctica. It was an approximately rectangular prism 160 km long, 40.0 km wide, and 250 m thick. (a) What is the mass of this iceberg, given that the density of ice is 917 kg/m^3? (b) How much heat transfer (in joules) is needed to melt it? (c) How many years would it take sunlight alone to melt ice this thick, if the ice absorbs an average of 100 W/m^2, 12.00 h per day?
a. 1.47 x 10^15 kg; b. 4.90 x 10^20 J; c. 48.5 y
(a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil releases 2.80 x 10^7 J of energy when burned. To illustrate this difficulty, calculate the number of liters of water that must be expended to absorb the energy released by burning 1.00 L of crude oil, if the water’s temperature rises from 20.0°C to 100°C, it boils, and the resulting steam’s temperature rises to 300°C at constant pressure. (b) Discuss additional complications caused by the fact that crude oil is less dense than water.
a. 9.35 L; b. Crude oil is less dense than water, so it floats on top of the water, thereby exposing it to the oxygen in the air, which it uses to burn. Also, if the water is under the oil, it is less able to absorb the heat generated by the oil.
To help prevent frost damage, 4.00 kg of water at 0°C is sprayed onto a fruit tree. (a) How much heat transfer occurs as the water freezes? (b) How much would the temperature of the 200-kg tree decrease if this amount of heat transferred from the tree? Take the specific heat to be 3.35 kJ/kg·°C, and assume that no phase change occurs in the tree.
a. 319 kcal; b. 2.00°C
A 0.0500-kg ice cube at -30.0°C is placed in 0.400 kg of 35.0°C water in a very well-insulated container. What is the final temperature?
First bring the ice up to 0°C and melt it with heat Q1: 4.74 kcal. This lowers the temperature of water by delta_T2: 23.15°C. Now, the heat lost by the hot water equals that gained by the cold water (T_f is the final temperature): 20.6°C
Indigenous people sometimes cook in watertight baskets by placing hot rocks into water to bring it to a boil. What mass of 500°C granite must be placed in 4.00 kg of 15.0°C water to bring its temperature to 100°C, if 0.0250 kg of water escapes as vapor from the initial sizzle? You may neglect the effects of the surroundings.
Let the subscripts r, e, v, and w represent rock, equilibrium, vapor, and water, respectively. m_r c_r (T_1 - T_e) = m_v L_v + m_w c_w (T_e - T_2); m_r = (m_v L_v + m_w c_w (T_e - T_2)) / c_r (T_1 - T_e) = (0.0250 kg) (2256 x 10^3 J/kg) + (3.975 kg) (4186 x 10^3 J/kg·°C) (100°C - 15°C) / (840 J/kg·°C) (500°C - 100°C) = 4.38 kg
(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The walls’ surface area is 120 m^2 and their inside surface is at 18.0°C, while their outside surface is at 5.00°C. (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?
a. 1.01 x 10^3 W; b. One 1-kilowatt room heater is needed.
Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is 37.0°C, the skin temperature is 34.0°C, the thickness of the fatty tissues between the core and the skin averages 1.00 cm, and the surface area is 1.40 m^2.
84.0 W
A man consumes 3000 kcal of food in one day, converting most of it to thermal energy to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?
2.59 kg
Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of 10.0 m^2 and a thermal conductivity twice that of glass wool with the rate of heat conduction through a 0.750-cm-thick window that has an area of 2.00 m^2, assuming the same temperature difference across each.
Q / t = k A (T_2 - T_1) / d, so that (Q / t)_wall / (Q / t)_window = k_wall A_wall d_window / k_window A_window d_wall = (2 x 0.042 J/s·m·°C) (10.0 m^2) (0.750 x 10^-2 m) / (0.84 J/s·m·°C) (2.00 m^2) (13.0 x 10^-2 m). This gives 0.0288 wall: window, or 35:1 window: wall.
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through area 14 cm radius and at a rate of 2256 W, what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick.
Q / t = k A (T_2 - T_1) / d = k A delta_T / d => delta_T = d (Q / t) / k A = (6.00 x 10^-3 m) (2256 W) / (0.84 J/s·m·°C) (1.54 x 10^-2 m^2) = 1046°C = 1.05 x 10^3 K
What is the percent error of thinking the melting point of tungsten is 3695°C instead of the correct value of 3695 K?
7.39%
How much stress is created in a steel beam if its temperature changes from -15°C to 40°C but it cannot expand? For steel, the Young’s modulus Y = 210 x 10^9 N/m^2. (Ignore the change in area resulting from the expansion.) (alpha_steel = 12 x 10^-6 /°C)
F / A = (210 x 10^9 Pa) (12 x 10^-6 /°C) (40°C - (-15°C)) = 1.4 x 10^8 N/m^2
A mercury thermometer still in use for meteorology has a bulb with a volume of 0.780 cm^3 and a tube for the mercury to expand into of inside diameter 0.130 mm. (a) Neglecting the thermal expansion of the glass, what is the spacing between marks 1°C apart? (b) If the thermometer is made of ordinary glass (not a good idea), what is the spacing? (beta_mercury = 180 x 10^-6 /°C, alpha_glass = 9 x 10^-6 /°C)
a. 1.06 cm; b. 1.11 cm
A 0.800-kg iron cylinder at a temperature of 1.00 x 10^3°C is dropped into an insulated chest of 1.00 kg of ice at its melting point. What is the final temperature, and how much ice has melted?
6.3°C. All of the ice melted.
Repeat the preceding problem with 0.500 kg of ice, assuming that the ice is initially in a copper container of mass 1.50 kg in equilibrium with the ice.
63.9°C, all the ice melted
a) Calculate the rate of heat conduction through a double-paned window that has a 1.50-m^2 area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside surface temperature is 15.0°C, while that on the outside is -10.0°C. (Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).
a. 83 W; b. 1.97 x 10^3 W; The single-pane window has a rate of heat conduction equal to 1969 / 83, or 24 times that of a double-pane window.
For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the difference in temperature is 2.00°C, and the skin area is 1.50 m^2. How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.)
The rate of heat transfer by conduction is 20.0 W. On a daily basis, this is 1,728 kJ/day. Daily food intake is 2400 kcal/d x 4186 J/kcal = 10,050 kJ/day. So only 17.2% of energy intake goes as heat transfer by conduction to the environment at this delta_T.
In a calorimeter of negligible heat capacity, 200 g of steam at 150°C and 100 g of ice at -40°C are mixed. The pressure is maintained at 1 atm. What is the final temperature, and how much steam, ice, and water are present?
The amount of heat to melt the ice and raise it to 100°C is not enough to condense the steam, but it is more than enough to lower the steam’s temperature by 50°C, so the final state will consist of steam and liquid water in equilibrium, and the final temperature is 100°C; 9.5 g of steam condenses, so the final state contains 49.5 g of steam and 40.5 g of liquid water.
An infrared heater for a sauna has a surface area of 0.050 m^2 and an emissivity of 0.84. What temperature must it run at if the required power is 360 W? Neglect the temperature of the environment.
620 K
A pendulum is made of a rod of length L and negligible mass, but capable of thermal expansion, and a weight of negligible size. (a) Show that when the temperature increases by dT, the period of the pendulum increases by a fraction alpha dT / 2. (b) A clock controlled by a brass pendulum keeps time correctly at 10°C. If the room temperature is 30°C, does the clock run faster or slower? What is its error in seconds per day? (alpha_brass = 19 x 10^-6 /°C)
Denoting the period by P, we know P = 2 pi sqrt(L / g). When the temperature increases by dT, the length increases by alpha L dT. Then the new length is a. P = 2 pi sqrt((L + alpha L dT) / g) = 2 pi sqrt(L / g (1 + alpha dT)) = 2 pi sqrt(L / g) (1 + 1/2 alpha dT) = P (1 + 1/2 alpha dT) by the binomial expansion. b. The clock runs slower, as its new period is 1.00019 s. It loses 16.4 s per day.
Find the growth of an ice layer as a function of time in a Dewar flask. Call the thickness of the ice layer L. (a) Derive an equation for dL/dt in terms of L, the temperature T above the ice, and the properties of ice (which you can leave in symbolic form instead of substituting the numbers). (b) Solve this differential equation assuming that at t = 0, you have L = 0. (c) Will the water eventually freeze to the bottom of the flask?
a. dL / dT = k T / rho L; b. L = sqrt(2 k T t / rho L_t); c. yes
