#$&* course PHY 202 2/21 about 2 pm 004. `query 4
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Given Solution: ** The change in the thermal energy of an isolated system is 0. So assuming that the systems are isolated the thermal energy change of one object is equal and opposite to that of the other. For an ideal substance the change in the thermal energy of an object is directly proportional to the change in its temperature and to its mass (i.e., more mass and more change in temperature implies more change in thermal energy). The specific heat is the proportionality constant for the substance. Using `dQ for the change in the thermal energy of an object we can express this as • `dQ = mass * specific heat * `dT. (General College and University Physics students note that most substances do not quite behave in this ideal fashion; for most substances the specific heat is not in fact strictly constant and for most substances changes with temperature.) For two objects combined in a closed system we have `dQ1 + `dQ2 = 0, which gives us the equation • m1 c1 `dT1 + m2 c2 `dT2 = 0 or equivalently • m1 c1 `dT1 = - m2 c2 `dT2. That is, whatever energy one substance loses, the other gains. In this situation we know the specific heat of water, the two temperature changes and the two masses. We can therefore solve this equation for specific heat c2 of the unknown substance. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin and gen problem 14.3: 2500 Cal per day is how many Joules per day? At a dime per kilowatt hour, how much would this cost? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: 1 Cal = 4186 J 2500 Cal/day * (4186 J / 1 Cal) = approx. 10,500,000 J/day 1 W = 1 J/s 1 kW = (1 J/s) * 1000 = 1000 J/s 1 kW/hr = 1000 J/s * 3600 s/hr = 3,600,000 J 10,500,000 J / 3,600,000 J /(kWh) = 2.92 kWh 2.92 kWh * 0.10 cents/kWh = 29.2 cents confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: One Cal (with capital C) is about 4200 Joules, so 2500 Cal is about 4200 * 2500 Joules = 10,500,000 Joules. A watt is a Joule per second, a kilowatt is 1000 Joules / second and a kiliowatt-hour is 1000 Joules / second * 3600 seconds = 3,600,000 Joules. 10,500,000 Joules / (3,600,000 Joules / kwh) = 3 kwh, rounded to the nearest whole kwh. This is about 30 cents worth of electricity, and a dime per kilowatt-hour. Relating this to your physiology: • You require daily food energy equivalent to 30 cents’ worth of electricity. • It's worth noting that you use 85% of the energy your metabolism produces just keeping yourself warm. • It follows that the total amount of physical work you can produce in a day is worth less than a dime. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: prin phy and gen phy problem 14.07 how many Kcal of thermal energy would be generated in the process of stopping a 1200 kg car from a speed of 100 km/hr? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: Since Joules are kg*m^2/s^2, the mass does not need converted but the velocity needs to be in m/s. 100 km/hr * (1000 m / km) * (1 hr / 3600 s) = 27.8 m/s KE = .5*m*v^2 = .5 * 1200 kg * (27.8 m/s)^2 = 463,704 J It will require 463,704 J to stop the car. 463,704 J * (1 Kcal / 4186 J) = 110.8 Kcal confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: NOTE: The given solution is based on student solutions for a previous edition of the text, in which the mass of the car was 1000 kg and its initial velocity 100 km/hr. The adjustments for the current mass and velocity (1200 kg and 95 km/hr as of the current edition) are easily made (see the student question below for a solution using these quantities). **STUDENT SOLUTION for 1000 kg car at 100 km/hr WITH ERROR IN CONVERSION OF km/hr TO m/s: The book tells that according to energy conservation • initial KE = final KE + heat or (Q) • 100km/hr *3600*1/1000 = 360 m/s INSTRUCTOR COMMENT: 100km/hr *3600 sec / hr *1/ (1000 m / km) = 360 km^2 sec / ( m hr^2), not 360 m/s. The correct conversion would be 100 km / hr is 100 * 1000 meters / (3600 sec) = 28 m/s or so. STUDENT SOLUTION WITH DIFFERENT ERROR IN UNITS Ke=0.5(1000Kg)(100Km)^2 = 5MJ 1Kcal=4186J 5MJ/4186J==1194Kcal INSTRUCTOR COMMENT: Right idea but 100 km is not a velocity and kg * km^2 does not give you Joules. 100 km / hr = 100,000 m / (3600 sec) = 28 m/s, approx. so KE = .5(1000 kg)(28 m/s)^2 = 400,000 J (approx.). or 100 Kcal (approx). ** STUDENT QUESTION: The book and this problem originally states *1200kg* as the mass. The solution uses 1000kg, giving the answer as about 100kcal. The book uses 95km/hr, 1200kg, and gets an answer of 100kcal. This problem shows 100km/hr, with a mass of 1000kg in the solution and still an answer of 100kcal. I just want to make sure I am doing it correctly. Where 26.39m/s should be used, I am using the conversion for this specific problem with 100km/hr = 1000m/1hr = 1hr/3600s = 27.78 ~28m/s. KE = 1/2mv^2 = ½(1200kg)(28 m/s)^2 = 470,400kg*m/s^2 = 470,400 J 470,400J = 1 cal/4.186J = 1 kcal/1000cal = 112.37 kcal = 112kcal INSTRUCTOR RESPONSE: I apparently missed the change in the latest edition of the text; or perhaps I chose not to edit the student solutions posted here. In any case your solution is good. Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: Openstax: 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 20.0ºC : (a) water; (b) concrete; (c) steel; and (d) mercury. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: ‘dQ = mc’dT or ‘dT = ‘dQ/(m*c) Mass for all is 1 kg and ‘dQ is 1 Kcal. ‘dT = 1Kcal / (1 kg * c), with c being the specific heat of each item and I will list below. Once ‘dT is determined, it must be added to the initial temperature of 20 C to determine final temp. a. water, c = 1 Kcal/(kg*C), ‘dT = 1 C, T_f = 21 C b. concrete, c = 0.20 Kcal/(kg*C), ‘dT = 5 C, T_f = 25 C c. steel, c = 0.108 Kcal/(kg*C), ‘dT = 9.3 C, T_f = 29.3 C d. mercury = 0.0333 Kcal/(kg*C), ‘dT = 30.03 C, T_f = 50 C confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The given solution will use 1 kcal / (kg C) as the specific heat of water, and will assume that the specific heat of concrete is about 1/3 this value, while that of steel is about 1/10 of this value. Your solution should used more accurate accepted values for the specific heats. 1 kcal of heat transferred into 1 kg of water will therefore raise its temperature by 1 Celsius. This is exactly what it means to say that the specific heat of water is 1 kcal / (kg C). The water will therefore end up at temperature 21 C. The specific heat of concrete being 1/3 that of water, the same heat transferred into 1 kg of concrete will raise its temperature 3 times as much, an increase of 3 Celsius, resulting in a final temperature of 23 C. Similar reasoning would conclude that the transfer of the same heat to 1 kg of steel would raise its temperature by 10 C, to 30 C. We could of course use the formula `dQ = m c `dT, solving for `dT to get `dT = `dQ / (m c). In each case m would be 1.00 kg, `dQ would be 1.00 kCal, and c would depend on the substance. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK #$&* ********************************************* Question: Openstax: A 0.250-kg block of a pure material is heated from 20.0ºC to 65.0ºC by the addition of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is most likely composed. The specific heat of .250 kg of material requiring 4.35 kJ to change its temperature by 45 Celsius is 4.35 kJ / (.250 kg * 45 C) = .4 kJ / (kg C), very approximately. Copper and zinc are two elements with a specific heat in this range. The given information is sufficient to calculate the specific heat to 3 significant figures, and this should be sufficient to determine the substance from a table of specific heats. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: M = 0.250 kg, ‘dT = 45 C, ‘dQ = 4.35 kJ = 4,350 J ‘dQ = mc’dT or C = ‘dQ/(m*’dT) = 4,350 J / (0.250 kg * 45 C) = 386.67 J/(kg*C) The specific heat of copper is listed as 387 J/(kg*C) so I believe the material is copper. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK #$&* ********************************************* Question: (optional for Principles of Physics) Suppose identical amounts of heat transfer into different masses of copper and water, causing identical changes in temperature. What is the ratio of the mass of copper to water? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: If all is the same except for specific heat, the equation could be solved as M = ‘dQ/(c*’dT), c_water = 1 and c_copper = 0.0924, units in (Kcal / kg*C), so 1 / 0.0924 = 10.8 times more mass of copper than water. confidence rating #$&*: 3 ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: The specific heat of copper is about 1/11 that of water, so 11 times the mass of copper would be required. This could also be formulated by an equation: m_1 c_1 `dT_1 = m_2 c_2 `dT_2. Since the temperature changes are identical, `dT_1 = `dT_2 so m_1 c_1 = m_2 c_2. Thus m_2 / m_1 = c_1 / c_2 and the ratio of copper mass to water mass is equal to the ratio of water's specific heat to that of copper, which is m_2 / m_1 = (4.18 J / (kg C) / (.386 J / (kg C) ), close to 1/11. &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique Rating:OK #$&* ********************************************* Question: query gen phy problem 14.13 .4 kg horseshoe into 1.35 L of water in .3 kg iron pot at 20 C, final temp 25 C, final temp. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘dQ1+’dQ2+’dQ3 = 0, where ‘dQ1 represents the iron horseshoe, ‘dQ2 represents the water and ‘dQ3 represents the iron pot. M1 = 0.4 kg, c1 = 452 J/(kg*C), ‘dT1 = ? M2 = 1.35 L * (1000 cm^3 / 1 L) * (1 g / cm^3) = 1350 g = 1.35 kg, c2 = 4186 J/(kg*C), ‘dT2 = 5 C M3 = 0.3 kg, c3 = 452 J/(kg*C), ‘dT3 = 5 C M1c1’dT1 + m2c2’dT2 + m3c3’dt3 = 0 (0.4 kg * 452 J/(kg*C) * ‘dT1) + (1.35 kg * 4186 J/(kg*C) * 5 C) + (0.3 kg * 452 J/(kg*C) * 5 C) = 0 180.8 J/C * ‘dT1 = -28933.5 J ‘dT1 = -160.03 C ‘dT1 = T1_f - T1_i -160.03 C = 25 C - T1_i T1_i = -(-160.03 C - 25 C) = 185.03 C confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
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Given Solution: ** STUDENT SOLUTION BY EQUATION (using previous version in which the amount of water was 1.6 liters; adjust for the present 1.35 liters): M1*C1*dT1 + M2*C2*dT2 = 0 is the equation that should be used. 0.4kg * 450J/Kg*Celcius *(25celcius -T1) + 0.3kg * 450J/kg*celcius * (25 celcius - 20 celcius) + 1.6kg *4186J/kg*celcius * (25 celcius - 20 celcius) = 0 Solve for T1, T1 = 214.8 Celsius Solution below is 189.8 C. GOOD STUDENT SOLUTION: This problem is very similar to the first video experiment. You can determine in the same way the thermal energy gained by the water. The pot gains a little thermal energy too,and you have to know how many Joules are gained by a kg of the iron per degree per kg, which is the specific heat of iron (give in text). You can therefore find the energy lost by the horseshoe, from which you can find the energy lost per degree per kg. For this problem I think we are assuming that none of the water is boiled off ass the hot horse shoe is dropped into the water. First I will find the initial temperature of the shoe. Since water's density is 1g/ml, 1 milliliter of water will weigh 1 gram. So 1.35 Liters of water will have a mass of 1.35 kg. 1.35kg of water is heated by 5 degrees • The specific heat of water is 4186 J/kg/degrees celsius so 4186 J / kg / C *1.35 kg * 5 C = 28255 J of energy is necessary to heat the water but since it is in equilibrium with the bucket it must be heated too. mass of bucket = 0.30 kg • specific heat of iron = 450 J/kg/degrees • 450 J / kg / C * 0.30 kg * 5 C = 675 J to heat bucket So it takes • 675 J to heat bucket to 25 degrees celsius • 28255 J to heat water to 25 degrees celsius so the horse shoe transferred 675+28255 = 28930 J of energy. Mass of horse shoe = 0.40 kg • horse shoe is also iron • specific heat of iron = 450 J/kg/degree • energy transferred / mass = 28930 J / 0.40kg =72,326 J / kg • 72 330 J / kg, at 450 (J / kg) / C, implies `dT = 72,330 J/kg / (450 J / kg / C) = 160.7 C, the initial temperature of the horseshoe. A symbolic solution: m1 c1 `dT1 + m2 c2 `dT2 + m3 c3 `dT3 = 0. Let object 1 be the water, object 2 the pot and object 3 the horseshoe. Then `dT1 = `dT2 = + 5 C, and `dT3 = 25 C - T_03, where T_03 is the initial temperature of the horseshoe. We easily solve for `dT3: `dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) so `dT3 = - (m1 c1 `dT1 + m2 c2 `dT2) / (m3 c3) = - (1.35 kg * 4200 J / (kg C) * 5 C + .3 kg * 450 J / (kg C) ) / ((.4 kg * 450 J / (kg C) ) = -160 C, approx. so 25 C - T_03 = -160 C and • T_03 = 160 C + 25 C = 185 C, approx.. STUDENT ERROR: MISSING COMMON KNOWLEDGE: Estimate 1.60L of water = 1KG. Could not find a conversion for this anywhere. INSTRUCTOR RESPONSE: Each of the following should be common knowledge: • 1 liter = 1000 mL or 1000 cm^3. • Density of water is 1 gram/cm^3 so 1 liter contains 1000 g = 1 kg. • Alternatively, density of water is 1000 kg / m^3 and 1 liter = .001 m^3, leading to same conclusion. ** Your Self-Critique:OK Your Self-Critique Rating:OK ********************************************* Question: `q001. Which requires more energy, a 100 kg person climbing a hill 200 meters high or a cup of water heated from room temperature to the boiling point? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: ‘dKE=-‘dPE ‘dPE = m*g*h = 100 kg * 9.8 m/s^2 * 200 m = 196,000 J ‘dQ = mc’dT, mass of 8 ounces of water = 8 ounces * 29.57 mL/1ounce = 236.56 mL, 237 g = 0.237 kg ‘dQ = 0.237 kg * 4186 J/kg*C * (100 C - 22 C) = approx. 77,200 J It requires more energy to climb the hill. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating:OK ********************************************* Question: `q002. A container holds 4 kilograms of water, 8 kilograms of concrete and 500 grams of ice, all at 0 Celsius. The system is heated to 30 Celsius. Place in order the heat required to raise the temperature of the these three components of the system, given common knowledge and the fact that the specific heat of concrete is about 1/3 that of water. At about what temperature would two of these components have gained the same amount of heat? YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your Solution: The mass of the concrete is twice the water but 1/3 the specific heat. The ice mass is 1/8 that of water and specific heat is about 1/2. Water would require the most heat. The concrete would be next since 2*(1/3) is greater than (1/8)*(1/3). Another way to look at it is multiplying the mass and the specific heat and get the heat required for one degree C. This puts water at 16744 J/C, concrete 6720 J/C, and ice 1045 J/C. The system though, would not change until the ice melted, changing it to water, and then the temperature would begin to rise. I guess until the temperature began to rise, that is when the ice melted, the concrete and water would gain the same amount of heat. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^ ------------------------------------------------ Self-Critique Rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!