How much work in Joules is required to pump all of the water over the side?
Q. A circular pool has a diameter of 16m, the sides are 4m high and the depth of the water is 2.5m. How much work in Joules is required to pump all of the water over the side? The acceleration due to gravity is = 9.8m/s^2 Density of water = 1000kg/m^3 Any hints on how to set up this problem? Thank you!
Asked by Mrs.Sizemore - Fri Apr 6 14:18:06 2007 - - 2 Answers - 0 Comments
A. There are several ways to do this, but I believe the most elegant would be so: 1. Find the center of gravity of all the water's mass (which would be the geometrical center of the water, at the middle of the water's depth). 2. Consider all the water to be at that point: calculate its total mass, and place it at that point. 3. The energy needed to dump the water is the energy needed to lift that mass to the edge of the pool (E = mgh, where m is its mass, g is gravity and h is the height to the edge). Try it yourself. If you get stuck, here's the detail: 1. The center of gravity of the water is at the middle of the circle and 1.25 m over the bottom. 2. The total mass of the water is its density (1000) times its volume (which is the… [cont.]
Answered by Pedro Gomez-Esteban - Fri Apr 6 14:36:14 2007
Q. A circular pool has a diameter of 16m, the sides are 4m high and the depth of the water is 2.5m. How much work in Joules is required to pump all of the water over the side? The acceleration due to gravity is = 9.8m/s^2 Density of water = 1000kg/m^3 Any hints on how to set up this problem? Thank you!
Asked by Mrs.Sizemore - Fri Apr 6 14:18:06 2007 - - 2 Answers - 0 Comments
A. There are several ways to do this, but I believe the most elegant would be so: 1. Find the center of gravity of all the water's mass (which would be the geometrical center of the water, at the middle of the water's depth). 2. Consider all the water to be at that point: calculate its total mass, and place it at that point. 3. The energy needed to dump the water is the energy needed to lift that mass to the edge of the pool (E = mgh, where m is its mass, g is gravity and h is the height to the edge). Try it yourself. If you get stuck, here's the detail: 1. The center of gravity of the water is at the middle of the circle and 1.25 m over the bottom. 2. The total mass of the water is its density (1000) times its volume (which is the… [cont.]
Answered by Pedro Gomez-Esteban - Fri Apr 6 14:36:14 2007
How many joules of solar energy fall on your back as you work on your tan for 0.70h?
Q. The intensity of electromagnetic waves from the sun is 1.4 kW/m^2 just above the earth's atmosphere. Eighty percent of this reaches the surface at noon on a clear summer day. Suppose you model your back as a 27cm & 51cm rectangle. How many joules of solar energy fall on your back as you work on your tan for 0.70h?
Asked by ginger_jj - Tue Apr 15 15:04:38 2008 - - 1 Answers - 1 Comments
A. 80% of 1.4 = 1.12 kW / m Area = 0.27 * 0.51 = 0.138 m Power = 1.12 * 0.138 = 0.154 kW Energy = Power * Time Time = 0.7 * 3600 = 2520 seconds Energy = 0.154 * 2520 = 389 Joules
Answered by WhatGoes Around Comes Around - Fri Apr 18 05:33:59 2008
Q. The intensity of electromagnetic waves from the sun is 1.4 kW/m^2 just above the earth's atmosphere. Eighty percent of this reaches the surface at noon on a clear summer day. Suppose you model your back as a 27cm & 51cm rectangle. How many joules of solar energy fall on your back as you work on your tan for 0.70h?
Asked by ginger_jj - Tue Apr 15 15:04:38 2008 - - 1 Answers - 1 Comments
A. 80% of 1.4 = 1.12 kW / m Area = 0.27 * 0.51 = 0.138 m Power = 1.12 * 0.138 = 0.154 kW Energy = Power * Time Time = 0.7 * 3600 = 2520 seconds Energy = 0.154 * 2520 = 389 Joules
Answered by WhatGoes Around Comes Around - Fri Apr 18 05:33:59 2008
How many joules of energy are stored up in one pound of fat?
Q. If a person was lifting weights, how many joules of work would a person have to do in order to burn off one pound of fat? Thanks
Asked by STL Biker - Fri Jan 4 21:26:19 2008 - - 3 Answers - 0 Comments
A. Per Wikipedia, "Energy intake to the body that is not used up is mostly stored as fat in the fat tissue. Consider the following theoretical calculation: About 3,500 kcal are contained in 1 pound of fat. If you eat 3,500 kcal more than your body needs, you will put on about 1 pound of fat. If you burn 3,500 kcal more than you eat, you will lose about 1 pound of fat. This assumes that all the weight gained and lost is in the form of fat. In reality, muscle and organ mass will change as well. However, this calculation is somewhat theoretical, as the exact increase in fat tissue also depends on the form of energy intake. For example, food consisting mainly of fat can be converted into fat tissue quite efficiently, whereas carbohydrates cannot." [cont.]
Answered by Philip J - Fri Jan 4 22:06:36 2008
Q. If a person was lifting weights, how many joules of work would a person have to do in order to burn off one pound of fat? Thanks
Asked by STL Biker - Fri Jan 4 21:26:19 2008 - - 3 Answers - 0 Comments
A. Per Wikipedia, "Energy intake to the body that is not used up is mostly stored as fat in the fat tissue. Consider the following theoretical calculation: About 3,500 kcal are contained in 1 pound of fat. If you eat 3,500 kcal more than your body needs, you will put on about 1 pound of fat. If you burn 3,500 kcal more than you eat, you will lose about 1 pound of fat. This assumes that all the weight gained and lost is in the form of fat. In reality, muscle and organ mass will change as well. However, this calculation is somewhat theoretical, as the exact increase in fat tissue also depends on the form of energy intake. For example, food consisting mainly of fat can be converted into fat tissue quite efficiently, whereas carbohydrates cannot." [cont.]
Answered by Philip J - Fri Jan 4 22:06:36 2008
How many joules of heat energy are needed to completely change 21.0 g of water to steam at the boiling tempera
Q. How many joules of heat energy are needed to completely change 21.0 g of water to steam at the boiling temperature?
Asked by Jacky - Wed May 21 23:14:36 2008 - - 1 Answers - 0 Comments
A. Heat of vaporization of water is 2270 J/g. 21.0 g x (2270 J / 1 g) = 47,700 J = 47.7 kJ
Answered by Chem Man - Wed May 21 23:21:12 2008
Q. How many joules of heat energy are needed to completely change 21.0 g of water to steam at the boiling temperature?
Asked by Jacky - Wed May 21 23:14:36 2008 - - 1 Answers - 0 Comments
A. Heat of vaporization of water is 2270 J/g. 21.0 g x (2270 J / 1 g) = 47,700 J = 47.7 kJ
Answered by Chem Man - Wed May 21 23:21:12 2008
How do you get a kinetic energy in joules and calories?
Q. Like if your given a molar mass and temperature, what formula can you use so you can get values of the kinetic energy into joules and calories. The only formula I can think of is root((3RT)/M) but that gives meters per second.
Asked by Peter P - Mon Feb 5 17:18:34 2007 - - 1 Answers - 0 Comments
A. Both Joules and Calories are units of energy. Joule-unit is associated with mechanical energy and Calorie with heat. 1 calories is equal to 4.184 Joules so the is .239 calories ories per Joule What you are thinking about is internal energy of a gas Ei=(3/2)nRT Then R is the ideal gas constant (joule per Kelvin per mole) n- moles of gas T degrees Kelvin And Ei- is energy in joules.
Answered by Edward - Mon Feb 5 18:03:01 2007
Q. Like if your given a molar mass and temperature, what formula can you use so you can get values of the kinetic energy into joules and calories. The only formula I can think of is root((3RT)/M) but that gives meters per second.
Asked by Peter P - Mon Feb 5 17:18:34 2007 - - 1 Answers - 0 Comments
A. Both Joules and Calories are units of energy. Joule-unit is associated with mechanical energy and Calorie with heat. 1 calories is equal to 4.184 Joules so the is .239 calories ories per Joule What you are thinking about is internal energy of a gas Ei=(3/2)nRT Then R is the ideal gas constant (joule per Kelvin per mole) n- moles of gas T degrees Kelvin And Ei- is energy in joules.
Answered by Edward - Mon Feb 5 18:03:01 2007
How much energy (in Joules) does a 222-hp engine expend when running for 29.2 minutes? ?
Q. How much energy (in Joules) does a 222-hp engine expend when running for 29.2 minutes?
Asked by Nolan S - Mon Oct 27 00:03:27 2008 - - 1 Answers - 0 Comments
A. first you need to convert the hp to watts and minutes to seconds since watt is J/sec 222 hp(746 watt/ 1 hp)= 165612 watt 29.2 min(60 sec/ 1 min)= 1752 sec then the definition of power P=E/t E=Pt=(165612 watt)(1752 sec) E=290152224 joules= 2.90x10^8 joules if you're watching for significant figures.
Answered by unknown - Mon Oct 27 00:18:54 2008
Q. How much energy (in Joules) does a 222-hp engine expend when running for 29.2 minutes?
Asked by Nolan S - Mon Oct 27 00:03:27 2008 - - 1 Answers - 0 Comments
A. first you need to convert the hp to watts and minutes to seconds since watt is J/sec 222 hp(746 watt/ 1 hp)= 165612 watt 29.2 min(60 sec/ 1 min)= 1752 sec then the definition of power P=E/t E=Pt=(165612 watt)(1752 sec) E=290152224 joules= 2.90x10^8 joules if you're watching for significant figures.
Answered by unknown - Mon Oct 27 00:18:54 2008
How can I convert joules to kilojoules?
Q. I'm learning a new topic in one of my classes that involves joules and so on and I would really appreciate your help. Please help. Thanks!
Asked by Sweety01 - Tue Dec 5 18:16:32 2006 - - 6 Answers - 0 Comments
A. 1 kilojoules = 10^3 joules if you see the word "kilo" in front of any unit, it will be 10^3 times as many as the base unit 1 kilogram = 10^3 grams
Answered by James Chan - Tue Dec 5 18:21:43 2006
Q. I'm learning a new topic in one of my classes that involves joules and so on and I would really appreciate your help. Please help. Thanks!
Asked by Sweety01 - Tue Dec 5 18:16:32 2006 - - 6 Answers - 0 Comments
A. 1 kilojoules = 10^3 joules if you see the word "kilo" in front of any unit, it will be 10^3 times as many as the base unit 1 kilogram = 10^3 grams
Answered by James Chan - Tue Dec 5 18:21:43 2006
What are the joules and kilojoules for the periodic element chromium?
Q. I have to do a project on the element chromium and for some reason i found everything but can't find the joules and kilo joules. Can anyone help? plz
Asked by Jameka Bing - Sat Feb 28 14:27:24 2009 - - 1 Answers - 0 Comments
Q. I have to do a project on the element chromium and for some reason i found everything but can't find the joules and kilo joules. Can anyone help? plz
Asked by Jameka Bing - Sat Feb 28 14:27:24 2009 - - 1 Answers - 0 Comments
What quantity of thermal energy (in Joules) is required to raise the temperature.. to it's melting point?
Q. What quantity of thermal energy (in Joules) is required to raise the temperature of 45 g of tin from 34.0 C, to its melting point, 231.9 C, and then melt the tin at that temperature? The specific heat of tin is 0.277 J/g.K, and the heat of fusion of this metal is 59.2 J/g.
Asked by hardluckwoman - Thu Mar 6 07:46:58 2008 - - 2 Answers - 0 Comments
A. Q = .227 delta T = 197.9 Lf = 59.2 so energy = Q*M*deltaT + Lf*M =(.227*45*197.9) + (59.2*45) =4686J (to nearest J)
Answered by troutsniff - Thu Mar 6 07:53:50 2008
Q. What quantity of thermal energy (in Joules) is required to raise the temperature of 45 g of tin from 34.0 C, to its melting point, 231.9 C, and then melt the tin at that temperature? The specific heat of tin is 0.277 J/g.K, and the heat of fusion of this metal is 59.2 J/g.
Asked by hardluckwoman - Thu Mar 6 07:46:58 2008 - - 2 Answers - 0 Comments
A. Q = .227 delta T = 197.9 Lf = 59.2 so energy = Q*M*deltaT + Lf*M =(.227*45*197.9) + (59.2*45) =4686J (to nearest J)
Answered by troutsniff - Thu Mar 6 07:53:50 2008
How to calculate kinetic energy in joules?
Q. Homework question. "A pendulum with a weight of 2 kg, initially at rest, swings from a position where its gravitational potential energy is 500 joules. How much kinetic energy does it possess at the bottom of its swing?" How do I do this?
Asked by Lily B - Tue Dec 30 16:12:27 2008 - - 1 Answers - 0 Comments
A. At the bottom of the swing, the pendulum has zero potential energy; it only has kinetic energy. That is, the original state of 500 joules potential energy has all been converted to kinetic energy of 500 joules.
Answered by capwest5a - Tue Dec 30 19:22:58 2008
Q. Homework question. "A pendulum with a weight of 2 kg, initially at rest, swings from a position where its gravitational potential energy is 500 joules. How much kinetic energy does it possess at the bottom of its swing?" How do I do this?
Asked by Lily B - Tue Dec 30 16:12:27 2008 - - 1 Answers - 0 Comments
A. At the bottom of the swing, the pendulum has zero potential energy; it only has kinetic energy. That is, the original state of 500 joules potential energy has all been converted to kinetic energy of 500 joules.
Answered by capwest5a - Tue Dec 30 19:22:58 2008
How would I find horsepower and joules of energy?
Q. A golf cart operates on a 24 V battery that provides 65 A for 4200 s and still being recharged like new. The golf cart motor draws current at 22 A. 1. What is the horsepower equivalent of the motor if 1 kW=1.34 horsepower? 2. How many joules of energy are available without hurting the battery? I know the total charge that the battery can provide without hurting it and also the total can that the cart can be driven between recharging. An equation would be very helpful with the given figures I gave. Thank you...
Asked by Rachelina - Sun Dec 2 21:03:03 2007 - - 1 Answers - 0 Comments
A. Volts times amps= Watts. 24*65= 1.595 kW 1.19 Horsepower. 2. After the movie.
Answered by Wylie Coyote - Sun Dec 2 22:05:11 2007
Q. A golf cart operates on a 24 V battery that provides 65 A for 4200 s and still being recharged like new. The golf cart motor draws current at 22 A. 1. What is the horsepower equivalent of the motor if 1 kW=1.34 horsepower? 2. How many joules of energy are available without hurting the battery? I know the total charge that the battery can provide without hurting it and also the total can that the cart can be driven between recharging. An equation would be very helpful with the given figures I gave. Thank you...
Asked by Rachelina - Sun Dec 2 21:03:03 2007 - - 1 Answers - 0 Comments
A. Volts times amps= Watts. 24*65= 1.595 kW 1.19 Horsepower. 2. After the movie.
Answered by Wylie Coyote - Sun Dec 2 22:05:11 2007
How much joules does it take to compress a spring?
Q. It takes 40N to compress a certain spring .8cm. How much in Joules does it take to compress the spring from 4cm to 5cm past its equilibrium? Please show your work. Thanks
Asked by Slick - Sat Feb 28 20:56:43 2009 - - 2 Answers - 0 Comments
A. Spring constant(k) = 40 / 8 . 10^ 3 = 5000 N/m Energy to compress 4 cm = 0.5 . k . x^2 = 0.5 . 5000 . (0.04)^2 = 4 J Energy to compress 5 cm = 0.5 . 5000 . (0.05)^2 = 6.25 J Energy needed to compress from 4 to 5 = 6.25 - 4 = 2.25 J Answer should be 2 J as only a single sig fig is supported by the data.
Answered by rex - Sun Mar 1 22:38:49 2009
Q. It takes 40N to compress a certain spring .8cm. How much in Joules does it take to compress the spring from 4cm to 5cm past its equilibrium? Please show your work. Thanks
Asked by Slick - Sat Feb 28 20:56:43 2009 - - 2 Answers - 0 Comments
A. Spring constant(k) = 40 / 8 . 10^ 3 = 5000 N/m Energy to compress 4 cm = 0.5 . k . x^2 = 0.5 . 5000 . (0.04)^2 = 4 J Energy to compress 5 cm = 0.5 . 5000 . (0.05)^2 = 6.25 J Energy needed to compress from 4 to 5 = 6.25 - 4 = 2.25 J Answer should be 2 J as only a single sig fig is supported by the data.
Answered by rex - Sun Mar 1 22:38:49 2009
How many joules of heat energy must be supplied?
Q. A home swimming pool contains 145 m3 of water. At the beginning of the swimming season, the water must be heated from 17 to 27 C. How many joules of heat energy must be supplied?
Asked by anthony - Thu Oct 22 16:52:54 2009 - - 1 Answers - 0 Comments
A. specific heat of water is 4.186 kJ/kgC 1 liter = 0.001 m = 0.998 kg at 20C 145 m of water x 0.998 kg/0.001 m = 145000 kg E = 4.186 kJ/kgC x 145000 kg x (27-17)K = 6.06e9 J = 6.06 GJ .
Answered by billrussell42 - Thu Oct 22 17:01:07 2009
Q. A home swimming pool contains 145 m3 of water. At the beginning of the swimming season, the water must be heated from 17 to 27 C. How many joules of heat energy must be supplied?
Asked by anthony - Thu Oct 22 16:52:54 2009 - - 1 Answers - 0 Comments
A. specific heat of water is 4.186 kJ/kgC 1 liter = 0.001 m = 0.998 kg at 20C 145 m of water x 0.998 kg/0.001 m = 145000 kg E = 4.186 kJ/kgC x 145000 kg x (27-17)K = 6.06e9 J = 6.06 GJ .
Answered by billrussell42 - Thu Oct 22 17:01:07 2009
How do I work out how many joules there are in food?
Q. How do I work out how many joules there are in food. I have burnt the food on a bunsen and used it to heat up the water in a tube and measured the temperature rise. I know the formula is something like 4.2Joules for every 1cmcubed of water and 1degree celsius heat change. Could someone confirm. So, I have 20cm3 water and a heat rise of 5degrees occurs, I would therefore have 1.05 JOULES
Asked by richard j - Mon Oct 1 14:04:24 2007 - - 5 Answers - 0 Comments
A. 1 cal = 4.1868 J Just take how many calories are in your food and multiply it by 4.1868 and you will have Joules. Hope this helps
Answered by aero - Mon Oct 1 14:11:40 2007
Q. How do I work out how many joules there are in food. I have burnt the food on a bunsen and used it to heat up the water in a tube and measured the temperature rise. I know the formula is something like 4.2Joules for every 1cmcubed of water and 1degree celsius heat change. Could someone confirm. So, I have 20cm3 water and a heat rise of 5degrees occurs, I would therefore have 1.05 JOULES
Asked by richard j - Mon Oct 1 14:04:24 2007 - - 5 Answers - 0 Comments
A. 1 cal = 4.1868 J Just take how many calories are in your food and multiply it by 4.1868 and you will have Joules. Hope this helps
Answered by aero - Mon Oct 1 14:11:40 2007
What is the final temperature (in joules) of the water and copper?
Q. A calorimeter containing 125 g of water at 25.8 degrees Celsius has added to it 105 g of copper at 112 degrees Celsius. Assuming no heat is lost, what is the final temperature of the water and copper (in joules)?
Asked by Jillbee - Tue Jan 6 02:08:33 2009 - - 1 Answers - 0 Comments
A. First: The final temperature cannot be expressed in Joules. Second: I will work on the basis that the calorimeter is not involved in the equation. Energy gained by water = energy lost by copper Let the final temperature be X Energy gained by water = mass * specific heat * change in temp. Energy gained H2O = 125 * 4.1813* (X - 25.8) Energy gained = 522.66(X -25.8) Energy gained = 522.66X - 13,484.7 Energy lost by Cu = 105*0.385*(112-X) Energy lost by Cu = 40.43 (112 - X) Energy lost by Cu = 4527.6 - 40.43X Equate these two equations 522.66X -13484.7 = 4527.6 - 40.43X 563.09X = 18,012.3 X = 31.988 Final temperature will be 32 C Note: I have not done a problem like this for perhaps 50 years. In my time, we took the calorimeter into… [cont.]
Answered by Trevor H - Tue Jan 6 03:42:22 2009
Q. A calorimeter containing 125 g of water at 25.8 degrees Celsius has added to it 105 g of copper at 112 degrees Celsius. Assuming no heat is lost, what is the final temperature of the water and copper (in joules)?
Asked by Jillbee - Tue Jan 6 02:08:33 2009 - - 1 Answers - 0 Comments
A. First: The final temperature cannot be expressed in Joules. Second: I will work on the basis that the calorimeter is not involved in the equation. Energy gained by water = energy lost by copper Let the final temperature be X Energy gained by water = mass * specific heat * change in temp. Energy gained H2O = 125 * 4.1813* (X - 25.8) Energy gained = 522.66(X -25.8) Energy gained = 522.66X - 13,484.7 Energy lost by Cu = 105*0.385*(112-X) Energy lost by Cu = 40.43 (112 - X) Energy lost by Cu = 4527.6 - 40.43X Equate these two equations 522.66X -13484.7 = 4527.6 - 40.43X 563.09X = 18,012.3 X = 31.988 Final temperature will be 32 C Note: I have not done a problem like this for perhaps 50 years. In my time, we took the calorimeter into… [cont.]
Answered by Trevor H - Tue Jan 6 03:42:22 2009
Calculate the quantity of work, in joules, associated with the compression?
Q. Calculate the quantity of work, in joules, associated with the compression of a gas from 5.73 L to 3.76 L by a constant pressure of 1.20 atm.
Asked by kym h - Thu Oct 25 22:24:42 2007 - - 1 Answers - 0 Comments
A. The work is W = V - p dV V because p is constant W = p (V - V ) = 1.2atm 101325Pa/atm (0.00573m - 0.00378m ) = 239.5323J
Answered by schmiso - Sat Oct 27 05:50:44 2007
Q. Calculate the quantity of work, in joules, associated with the compression of a gas from 5.73 L to 3.76 L by a constant pressure of 1.20 atm.
Asked by kym h - Thu Oct 25 22:24:42 2007 - - 1 Answers - 0 Comments
A. The work is W = V - p dV V because p is constant W = p (V - V ) = 1.2atm 101325Pa/atm (0.00573m - 0.00378m ) = 239.5323J
Answered by schmiso - Sat Oct 27 05:50:44 2007
How many joules of energy will be absorbed to melt a 44.0 gram ice cube?
Q. How many joules of energy will be absorbed to melt a 44.0 gram ice cube? The heat of fusion for water is 334 J/g. 14700 J 0.132 J 7.59 J 816 J 265000 J What formula do I need to use for this?
Asked by Hula - Thu Jul 2 15:00:57 2009 - - 2 Answers - 0 Comments
A. mass heat of fusion The masses must be the same, for example if the heat of fusion of water is 6012 J/mol and you multiply it times 44g, that won't work because the units of mol and g won't cancel out. In which case you would divide 44g / (18g/mol) = 2.4443 mol 2.4443 mol 6012 J/mol = 14695 J 334 J/g 44 g = 14696 J or 14.7 kJ
Answered by Mauri - Thu Jul 2 15:35:24 2009
Q. How many joules of energy will be absorbed to melt a 44.0 gram ice cube? The heat of fusion for water is 334 J/g. 14700 J 0.132 J 7.59 J 816 J 265000 J What formula do I need to use for this?
Asked by Hula - Thu Jul 2 15:00:57 2009 - - 2 Answers - 0 Comments
A. mass heat of fusion The masses must be the same, for example if the heat of fusion of water is 6012 J/mol and you multiply it times 44g, that won't work because the units of mol and g won't cancel out. In which case you would divide 44g / (18g/mol) = 2.4443 mol 2.4443 mol 6012 J/mol = 14695 J 334 J/g 44 g = 14696 J or 14.7 kJ
Answered by Mauri - Thu Jul 2 15:35:24 2009
How many Joules are given off when 50 grams of water at 0 degrees freezes?
Q. Would it be 0 joules?
Asked by Passion - Sat Apr 5 01:35:52 2008 - - 1 Answers - 0 Comments
A. This has to do with the Latent Heat of the water. It's the amount of energy required for a phase change to occur. In this case, the phase change is from Liquid to a Solid. Q=mL Q=Amount of Joules [J] m=mass (50 g) L=Latent Heat of the substance (334 J/g for water) So Q=50g x 334 J/g Q=16,700 Joules are given off
Answered by Chase - Sat Apr 5 02:20:03 2008
Q. Would it be 0 joules?
Asked by Passion - Sat Apr 5 01:35:52 2008 - - 1 Answers - 0 Comments
A. This has to do with the Latent Heat of the water. It's the amount of energy required for a phase change to occur. In this case, the phase change is from Liquid to a Solid. Q=mL Q=Amount of Joules [J] m=mass (50 g) L=Latent Heat of the substance (334 J/g for water) So Q=50g x 334 J/g Q=16,700 Joules are given off
Answered by Chase - Sat Apr 5 02:20:03 2008
How much heat energy, in joules, is required to raise the temperature of 488 g of tin from room temperature, 2
Q. How much heat energy, in joules, is required to raise the temperature of 488 g of tin from room temperature, 24.0 C, to its melting point, 231.9 C, and then melt the tin at that temperature?___J The specific heat of tin is 0.227 J/g K, and the metal requires 59.2 J/g to convert the solid to a liquid.
Asked by beeboroachgoingon197 - Mon Oct 8 11:05:26 2007 - - 1 Answers - 0 Comments
A. This is a 2 step process. First determine the amount of heat needed to heat the tin from 24.0 C to 231.9 C, a difference of 207.9 K. H = 488 g * 207.9 K * 0.227 J/g K = 23000 J The next step is to determine the amount of heat needed to melt the tin. H = 488 g * 59.2 J/g = 28900 J Now add the 2 amounts together to get the total amount of heat required to heat and melt the tin.
Answered by Lance - Thu Oct 11 22:17:29 2007
Q. How much heat energy, in joules, is required to raise the temperature of 488 g of tin from room temperature, 24.0 C, to its melting point, 231.9 C, and then melt the tin at that temperature?___J The specific heat of tin is 0.227 J/g K, and the metal requires 59.2 J/g to convert the solid to a liquid.
Asked by beeboroachgoingon197 - Mon Oct 8 11:05:26 2007 - - 1 Answers - 0 Comments
A. This is a 2 step process. First determine the amount of heat needed to heat the tin from 24.0 C to 231.9 C, a difference of 207.9 K. H = 488 g * 207.9 K * 0.227 J/g K = 23000 J The next step is to determine the amount of heat needed to melt the tin. H = 488 g * 59.2 J/g = 28900 J Now add the 2 amounts together to get the total amount of heat required to heat and melt the tin.
Answered by Lance - Thu Oct 11 22:17:29 2007
How do you find the number of joules per mole?
Q. Its a question on a lab, -Find the number of Joules per mole for Part A. Divide your answer from (2) by (3) I dont understand!
Asked by kayy - Thu Apr 16 17:22:57 2009 - - 2 Answers - 0 Comments
A. I am pretty sure it's done using the formula Q=mcDt Q=joules m=mass in kg Dt(delta T) = Tf-Ti (final temperature minus initial temperature) Use a mole calculator to calculate moles to mass.
Answered by Heart Cooks Brain - Thu Apr 16 17:30:26 2009
Q. Its a question on a lab, -Find the number of Joules per mole for Part A. Divide your answer from (2) by (3) I dont understand!
Asked by kayy - Thu Apr 16 17:22:57 2009 - - 2 Answers - 0 Comments
A. I am pretty sure it's done using the formula Q=mcDt Q=joules m=mass in kg Dt(delta T) = Tf-Ti (final temperature minus initial temperature) Use a mole calculator to calculate moles to mass.
Answered by Heart Cooks Brain - Thu Apr 16 17:30:26 2009
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