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<press any key to continue> How much heat is required to raise the temperature of 68 g of aluminum fluoride from 25oC to 80oC? Before continuing on with this tutorial, you should copy down the above problem and do the best you can do to solve it. Once you have the calculation completed, you should then go through the tutorial and see how you performed, paying particular attention to any areas where you may have made a mistake. <press any key to continue> How much heat is required to raise the temperature of 68 g of aluminum fluoride from 25oC to 80oC? Whenever we have a problem which deals with HEAT, we must recall the following relationship…. HEATgained or lost = (Mass)(T)(Cp) Keep in mind that the mass must be in grams, the temperature in degrees Celsius, and the specific heat capacity in either cal/goC or J/goC. So such a problem has 4 variables -- Heat, Mass, T, Cp Looking at our problem, we see the following values for these variables…….. Heat = x Mass = 68 g T = 80o C - 25o C = 55o C Cp = ? <press any key to continue> How much heat is required to raise the temperature of 68 g of aluminum fluoride from 25oC to 80oC? HEATgained or lost = (Mass)(T)(Cp) Heat = x Mass = 68 g T = 55o C Cp = ? = 0.895 J/goC Since the problem does not give us the Cp for aluminum fluoride, we refer to our list of Common Specific Heats and find that the value is………... Cp AlF3 = 0.895 J/goC We can substitute all of our values into the equation and solve the problem……... x = (68 g)(55oC)(0.895 J/goC) x = 3347.3 J <press any key to continue> In order to make 4 cups of tea, 1000 g of water is heated from 22o C to 99o C. How much energy is required? Before continuing on with this tutorial, you should copy down the above problem and do the best you can do to solve it. Once you have the calculation completed, you should then go through the tutorial and see how you performed, paying particular attention to any areas where you may have made a mistake. <press any key to continue> In order to make 4 cups of tea, 1000 g of water is heated from 22o C to 99o C. How much energy is required? Whenever we have a problem which deals with HEAT, we must recall the following relationship…. HEATgained or lost = (Mass)(T)(Cp) Keep in mind that the mass must be in grams, the temperature in degrees Celsius, and the specific heat capacity in either cal/goC or J/goC. So such a problem has 4 variables -- Heat, Mass, T, Cp Looking at our problem, we see the following values for these variables…….. Heat = x Mass = 1000 g T = 99o C - 22o C = 77o C Cp = ? <press any key to continue> In order to make 4 cups of tea, 1000 g of water is heated from 22o C to 99o C. How much energy is required? HEATgained or lost = (Mass)(T)(Cp) Heat = x Mass = 1000 g T = 77o C Cp = ? = 1.0 cal/goC Since the problem does not give us the Cp for water, we refer to our list of Common Specific Heats and find that the value is………... Cp H2O = 1.0 cal/goC We can substitute all of our values into the equation and solve the problem……... x = (1000 g)(77oC)(1.0 cal/goC) x = 77,000 cal <press any key to continue> If a piece of gold with a mass of 45.5 g and a temperature of 80.5oC is dropped into 192 g of water at 15oC, what is the final temperature of the system? Before continuing on with this tutorial, you should copy down the above problem and do the best you can do to solve it. Once you have the calculation completed, you should then go through the tutorial and see how you performed, paying particular attention to any areas where you may have made a mistake. <press any key to continue> If a piece of gold with a mass of 45.5 g and a temperature of 80.5oC is dropped into 192 g of water at 15oC, what is the final temperature of the system? Whenever we have a problem which deals with HEAT, we must recall the following relationship…. HEATgained or lost = (Mass)(T)(Cp) Keep in mind that the mass must be in grams, the temperature in degrees Celsius, and the specific heat capacity in either cal/goC or J/goC. And since this problem involves TWO objects at DIFFERENT TEMPERATURES being combined, we must also recall…... HEATgained = HEATlost In other words, when we combine the two objects, the heat lost by the “warm” object will be gained by the “cool” object, or, put another way, the temperature of the “warm” object will decrease and the temperature of the “cool” object will increase. <press any key to continue> If a piece of gold with a mass of 45.5 g and a temperature of 80.5oC is dropped into 192 g of water at 15oC, what is the final temperature of the system? HEATgained or lost = (Mass)(T)(Cp) HEATgained = HEATlost The problem asks us to determine the final temperature, so we will use “x” to represent that temperature. Looking at our list of COMMON SPECIFIC HEATS, we see that the Cp for gold = 0.13 J/goC and Cp for water = 4.18 J/goC. We now can re-write HEATgained = HEATlost ……... (Masswater)(Twater)(Cp water) = (Massgold)(Tgold)(Cp gold) Now we fill in this equation with our data……... (192 g)(x - 15o C)(4.18 J/goC) = (45.5 g)(80.5o C - x)(0.13 J/goC) Lastly, we solve for x…... (802.56x - 12038.4) = (476.16 - 5.92x) 808.48x = 12514.56 x = 15.48o C The final temperature of the system will be 15.48o C. <press any key to continue> A piece of unknown metal with mass 14.9 g is heated to 100o C and dropped into 75 g of water at 20o C. The final temperature of the system is 28.5o C. What is the specific heat of the metal? Before continuing on with this tutorial, you should copy down the above problem and do the best you can do to solve it. Once you have the calculation completed, you should then go through the tutorial and see how you performed, paying particular attention to any areas where you may have made a mistake. <press any key to continue> A piece of unknown metal with mass 14.9 g is heated to 100o C and dropped into 75 g of water at 20o C. The final temperature of the system is 28.5o C. What is the specific heat of the metal? Whenever we have a problem which deals with HEAT, we must recall the following relationship…. HEATgained or lost = (Mass)(T)(Cp) Keep in mind that the mass must be in grams, the temperature in degrees Celsius, and the specific heat capacity in either cal/goC or J/goC. And since this problem involves TWO objects at DIFFERENT TEMPERATURES being combined, we must also recall…... HEATgained = HEATlost In other words, when we combine the two objects, the heat lost by the “warm” object will be gained by the “cool” object, or, put another way, the temperature of the “warm” object will decrease and the temperature of the “cool” object will increase. <press any key to continue> A piece of unknown metal with mass 14.9 g is heated to 100o C and dropped into 75 g of water at 20o C. The final temperature of the system is 28.5o C. What is the specific heat of the metal? HEATgained or lost = (Mass)(T)(Cp) HEATgained = HEATlost The problem asks us to determine the specific heat of the metal, so we will use “x” to represent that specific heat. Looking at our list of COMMON SPECIFIC HEATS, we see that the Cp for water = 4.18 J/goC. We now can re-write HEATgained = HEATlost ……... (Masswater)(Twater)(Cp water) = (Massmetal)(Tmetal)(Cp metal) Now we fill in this equation with our data……... (75 g)(28.5o C - 20o C)(4.18 J/goC) = (14.9 g)(100o C - 28.5o C)(x J/goC) Lastly, we solve for x…... (2664.75) = (1065.35x) x = 2.5 J/goC The specific heat of the metal is 2.5 J/goC.

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posted: | 10/4/2012 |

language: | English |

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