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### A 55.0g aluminum block initially at 27.5 C absorbs 725 J of ...

The final temperature of aluminum is 42.1 C. QmcT This means that the amount of energy produced is equal to the mass of the system multiplied by its change of temperature and multiplied by its specific heat. LET 39S PUT IN WHAT WE KNOW Q quot725 Jquot mquot55.0 gquot c quot0.900 J/ Cg quot T quotfinal temperature initial temperaturequot T x27.5 quot Cquot We solve for T. quot725 Jquot quot55.0 g ...

### Solved A 55.0g aluminum block initially at 27.5 degree C ...

85 40 ratings Mass of aluminum 55 g Initial tempe . View the full answer. Transcribed image text: A 55.0g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum

### A 55.0g aluminum block initially at 27.5 C absorbs 725 J of ...

Answer to: A 55.0g aluminum block initially at 27.5 C absorbs 725 J of heat. What is the final temperature of the aluminum The specific heat of...

### CHM118 Unit One Flashcards Quizlet

A 55.0g aluminum block initially at 27.5 C absorbs 725 J of heat. What is the final temperature of the aluminum TfTi T TfTi q/mCs

### Thermodynamics Yeah Chemistry

A 27.5 g aluminum block is warmed to 65.0 degrees C and plunged into an insulated beaker containing 55.4 g water initially at 22.1 degress C. The aluminum and the water are allowed to come to thermal equilibrium.

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A 55.0g aluminum block initially at 27.5 C absorbs 725 J of heat. ... A 32.5 g cube of aluminum initially at 45.8 C is submerged into 105.3 g of water at 15.4 ...

### Chapter 10 Worksheet 2 Answer

If 52.5 kJ of heat is heated to a 1.02 kg block of metal, the temperature of the metal increases by 11.2 o C. Calculate the specific heat capacity of the metal in J/g o C. c q / m t convert units 52.5 KJ 52,500 J and 1.02 kg 1,020 g

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17. A sample of oxygen O2 of mass 30.0 g is confined in a vessel at 0 oC and 1000. torr. Then, 8.00 g of hydrogen H2 is pumped into the vessel at constant temperature. What will be the final pressure in the vessel 30.0 g O2 1 mol O2 0.938 mol O2 8.00 g H2 1 mol H2 3.96 mol H2

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2. A 55.0 g aluminum block initially at 27.5 C absorbs 725 of heat. What is the final temperature of the aluminum Cs aluminum 0.900 J/g C 4 points Question: 2. A 55.0 g aluminum block initially at 27.5 C absorbs 725 of heat. What is the final temperature of the aluminum Cs aluminum 0.900 J/g C 4 points

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The specific heat of aluminum is 897 J/kg K. This value is almost 2.3 times of the specific heat of copper. You can use this value to estimate the energy required to heat a 500 g of aluminum by 5 C, i.e., Q m x Cp x T 0.5 897 5 2242.5 J.

### A 27.5 g aluminum block is warmed to 65.9 C and plunged ...

A 27.5 g aluminum block is warmed to 65.9 C and plunged into an insulated beaker containing 55.5 g water initially at 22.1 C. The aluminum and the water are allowed to come to thermal equilibrium.Assuming that no heat is lost, what is the final temperature of the water and aluminum

### Calculating Heat

temperature if a 200.0 g block of gold at 100.0 C is placed in a coffeecup calorimeter containing 50.0 g of water at an initial temperature of 25.0 C. 1. How many calories are needed to raise the temperature of exactly 500 g of water from 22.15 C to 24.70 C 2. How many joules are released when 30.0 mL of chloroform cool 18.0 C The s

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The specific heat of the meta is C p 0.189 J/ g^oC This is a thermoequillibrium situation. We can use the equation Loss of Heat of the Metal Gain of Heat by the Water Q m Q w Q mDeltaTC p Q Heat m mass DeltaT T fT i T f Final Temp T i Initial Temp C P Specific Heat Metal Water m 17.5 g T f30.0^oC T i125.0^oC C P x Water m 15.0 g T f30.0^oC T i25.0^oC C P 4 ...

### Suppose that 26 g of each of the following... Clutch Prep

FREE Expert Solution. q mc T. c Al 0.900 J/g C. Convert q in J: q 2. 40 kJ 10 3 J 1 kJ 2400 J. Calculate the final temperature Tf : 87 123 ratings Problem Details. Suppose that 26 g of each of the following substances is initially at 28.0 C.

### CHEM 1411 Chapter 12 Homework Answers

17. A sample of oxygen O2 of mass 30.0 g is confined in a vessel at 0 oC and 1000. torr. Then, 8.00 g of hydrogen H2 is pumped into the vessel at constant temperature. What will be the final pressure in the vessel 30.0 g O2 1 mol O2 0.938 mol O2 8.00 g H2 1 mol H2 3.96 mol H2

### The final temp after warm metal is put into colder water

Example 10: Find the mass of liquid H 2 O at 100.0 C that can be boiled into gaseous H 2 O at 100.0 C by a 130.0 g Al block at temp 402.0 C Assume the aluminum is capable of boiling the water until its temperature drops below 100.0 C. The heat capacity of aluminum is 0.900 J g 1 C 1 and the heat of vaporization of water at 100 ...

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Problem 4: A 5.00 g sample of aluminum specific heat capacity 0.89 J g 1 C 1 and a 10.00 g sample of iron specific heat capacity 0.45 J g 1 C 1 are heated to 100.0 C. The mixture of hot iron and aluminum is then dropped into 91.9 g of water at 23.7 C.

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0.387 J/g C : 2. When 435 J of heat is added to 3.4 g of olive oil at 21 C, the temperature increases to 85 C. What is the specific heat of the olive oil Answer: 435 J/ 3.4g x 64 C 2.0 J/g C: 3. A piece of stainless steel with a mass of 1.55 g absorbs 141 J of heat when its temperature increases by 178 C.

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temperature if a 200.0 g block of gold at 100.0 C is placed in a coffeecup calorimeter containing 50.0 g of water at an initial temperature of 25.0 C. 1. How many calories are needed to raise the temperature of exactly 500 g of water from 22.15 C to 24.70 C 2. How many joules are released when 30.0 mL of chloroform cool 18.0 C The s

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One mole of O atoms corresponds to 15.9994 g Two moles of H atoms corresponds to 2 x 1.0079 g Sum molar mass 18.0152 g H 2O per mole Chapter 3 Calculation of Molar Masses Calculate the molar mass of the following Magnesium nitrate, Mg NO 3 2 1 Mg 24.3050 2 N 2x 14.0067 28.0134 6 O 6 x 15.9994 95.9964 Molar mass of Mg NO 3 2 148 ...

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If 52.5 kJ of heat is heated to a 1.02 kg block of metal, the temperature of the metal increases by 11.2 o C. Calculate the specific heat capacity of the metal in J/g o C. c q / m t convert units 52.5 KJ 52,500 J and 1.02 kg 1,020 g

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1 The specific heat of aluminum is 0.900 J/g oC. How much heat is required to raise the temperature of a 30.0g block of aluminum from 25.0oC to 75.0oC a. 0.540 J Incorrect b. 1.50 J Incorrect c. 1350 J Correct d. 1670 J Incorrect 2 Given the balanced equation representing a reaction at 101.3 kPa and 298K: N 2 g 3H 2 g 2NH 3

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25 g of aluminum 2.70 g/mL 1 Plan: Calculate the volume for each metal and select the metal sample with the greatest volume. 1 25g x 1 mL 9.3 mL aluminum 2.70 g 2 45 g x 1 mL 2.3 mL gold 19.3 g 3 75 g x 1 mL 6.6 mL lead 11.3 g Ref: Timberlake, Chemistry, Pearson/Benjamin Cummings, 2006, 9th Ed.

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12. How many grams of aluminum will react with 12.0 L of oxygen at STP as shown below: 4 Al s 3 O 2 g 2 Al 2 O 3 s 2 2 2 1 mol O 4 mol Al27.0 g 12.0 L O x x x 19.3 g Al 22.4 L 3 mol O 1 ,ol 13.

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How many kilocalories per gram are there in a 5.00g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100kg aluminum cup, causing a 54.9C temperature increase b Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.

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183.6 kJ / 0.5 mole H 2 g 367.2 kJ per mole of H 2 g . A 1.5 kg block of Ni at 100 C is placed into 500 mL of water that has a temperature of 21 C. What is the final temperature assuming the specific heat of Ni is 0.44 J/g C and the specific heat of water is 4.184 J/g C.

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### A 55.0g aluminum block initially at 27.5 degree C absorbs ...

A 55.0g aluminum block initially at 27.5 degree C absorbs 725 J of heat. What is the final temperature of the aluminum Express

### A 27.5 g aluminum block is warmed to 65.6 C and plunged ...

A 27.5 g aluminum block is warmed to 65.6 C and plunged into an insulated beaker containing 55.5 g water initially at 22.0 C. The aluminum and the water are allowed to come to thermal equilibrium.

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In the last video, we had a ten kilogram mass sitting on top of an inclined plane at a 30 degree angle And in order to figure out what would happen to this block we broke down the force of gravity on this block into the components that are parallel to the surface of the plane and perpendicular to the surface of the plane and for a perpendicular component, we got 49 times the square root of 3 N ...

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Solution. Again, you use q mcT, except you assume q aluminum q water and solve for T, which is the final temperature. You need to look up the specific heat values c for aluminum and water. This solution uses 0.901 for aluminum and 4.18 for water: Helmenstine, Todd. quotCalculating the Final Temperature of a Reaction From Specific Heat ...

### Specific Heat Calculations Chemistry for NonMajors

A 15.0 g piece of cadmium metal absorbs 134 J of heat while rising from 24.0 C to 62.7 C. Calculate the specific heat of cadmium. Step 1: List the known quantities and plan the problem . Known. heat 134 J mass 15.0 g Unknown. The specific heat equation can be rearranged to solve for the specific heat. Step 2: Solve .

### Table of Specific Heats HyperPhysics Concepts

Substance: c in J/gm K: c in cal/gm K or Btu/lb F: Molar C J/mol K: Aluminum: 0.900: 0.215: 24.3: Bismuth: 0.123: 0.0294: 25.7: Copper: 0.386: 0.0923: 24.5: Brass: 0 ...

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An aluminum cylinder has a mass of 878 grams, a diameter of 3.8 cm, and a height of 23 cm. What is its density 9783 9 3 A perfect cube has a mass of 610 g and a density of 4.8 g/cc. What is the length of one of its sides cc . Ypy place2SO ml of water in it. A marble has a mass of 120 g in f you place the marble in it 10.

### Temperature Change and Heat Capacity Physics

How many kilocalories per gram are there in a 5.00g peanut if the energy from burning it is transferred to 0.500 kg of water held in a 0.100kg aluminum cup, causing a 54.9C temperature increase b Compare your answer to labeling information found on a package of peanuts and comment on whether the values are consistent.

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### A 55.0g aluminum block initially at 27.5 c absorbs 725 j of ...

A 55.0g aluminum block initially at 27.5 c absorbs 725 j of heat. what is the final temperature of the aluminum express your answer in degrees celsius to one decimal place.

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We then determine the heat lost for a 55 kg person as: q q m 17000 J kg1 55 kg 9.2 105 J person 4 K And finally the mass of rations that needs to be consumed is given by: 31.0 g 3.0 10 J g 9.2 10 J q q m 4 1 5 rations person rations

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string, and the accelerations of the blocks have the same magnitude. Note that we take the positive direction in the direction of the acceleration for each block. We write F ma from the force diagram for each block: ycomponent block 1 : F T m 1 g m a ycomponent block 2 : m 2 g F T m 2 a.

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A block of solid with a mass of 10 kg is heated from 25oC to 200oC. If the change in the specific internal energy is found to be 67.55 kJ/kg, identify the material. Manual Solution TEST Solution Answers: Copper. A block of aluminum with a mass of 10 kg is heated from 25oC to 200oC.

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A 27.5 aluminum block is warmed to 65.8 and plunged into an insulated beaker containing 55.3 water initially at 22.3. The aluminum and the water are allowed to come to thermal equilibrium. Assuming that no heat is lost, what is the final temperature of t...

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Similarly, PTP7S EENFLGALFKALSKLL also maintains an helical secondary structure and selfassembles to construct nanofiber aggregates 23, 55 . Finally, the C 16 G 7 ERGDS peptide displays a polyproline type II helical coil conformation when any of its glycines 57 is replaced by an alanine .

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In an experiment, 1.2 g of sodium hydroxide pellets, NaOH s , were dissolved in 100 mL of water at 25 C. The temperature of the water rose to 27.5 C. Calculate the enthalpy change heat of solution for the reaction in kJ mol1 of solute. Calculate the heat released, q, in joules J , by the reaction:

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H neut 55.90 kJ mol 1. because no bonds need to be broken, and because making the HO bonds in H 2 O releases energy breaking bonds is an endothermic process, making bonds is an exothermic process Less than 55.90 kJ mol1 of energy is released when: a a weak acid neutralises a strong base b a strong acid neutralises a weak base

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Determine the accumulated and the discounted value of 1000 over 55 days at 7 using both ordinary and exact simple interest. 3. A taxpayer expects an income tax refund of 380 on May 1.