A hot metal at 125 oC was dropped into 75.0 g of
water kept at a room temperature of 20.6 oC in a
calorimeter. Within a couple of minutes, water’s temperature
increased to 26.8 oC?
(a) How much heat energy was absorbed by water? [‘c’ for water is
4.184 J/goC]
(b) How much heat energy was released by the metal?
(c) What information is needed to calculate the specific heat
capacity of the metal?
A hot metal weighing 65.0 g was heated to 125 oC and dropped into 75.0 g of water kept at a room temperature of 20.6 oC in a calorimeter. Within a couple of minutes, water’s temperature increased to 26.8 oC and stayed constant at that temperature. If the heat capacity or calorimetry constant is 11.5 J/oC, find the specific heat capacity of the metal.
Consider the dissolution of 1.50 grams of salt XY in 75.0 mL of water within a calorimeter. The temperature of the water decreased by 0.93 oC. The heat capacity of the calorimeter is 42.2 J/oC. The density of the water (and the solution) is 1.00 g/mL. The specific heat capacity of the solution is 4.184 J/goC. Identify the surroundings. Group of answer choices Salt dissolving in the water Calorimeter Water Solution and calorimeter
structure.com/courses/1253444/quizzes/2442703/take Calculate the amount of heat water absorbs from a piece of hot metal using the following data: 75.0 g of cold water is placed in a calorimeter. The initial temperature of the water is 21.2°C. To the calorimeter a 29.458 g piece of metal at 98.9 °C is added. The final temperature of the contents of the calorimeter is measured to be 29.5°C. (HINT: the specific heat of water is 4.184 ) heat absorbed by water = Knowing the...
3. A75.0 g piece of copper metal is initially at 100°C. It is dropped into a coffee cup calorimeter containing 75.0 g of water a a rature of 20.0°c. Assuming that the only heat exchange is between the copper metal and the water (no heat is given to the calorimeter), what is the final temperature of the water. Specific heat of copper 0.387 J/goC
An 78.5 g piece of metal whose T = 63.00 oC is placed in a coffee cup calorimeter containing 125 g water. When the system reaches equilibrium, the water has changed from 20.00 oC to 24.00 oC. What is specific heat of metal? An 88.5 g piece of metal whose T = 78.8 oC is placed in a coffee cup calorimeter containing 244 g water. When the system reaches equilibrium, the water has changed from 18.80 oC to 200 oC....
An unknown metal sample of 54 g at 108°C is dropped into a calorimeter cup containing 190 g of 2. water at 20.3°c. After equilibrium is reached, the temperature of water is increased to 24.6°C. What is the specific heat of this metal? identify the metal by looking up the specific heat table. (Specific heat of water is 4186 J/kg.°C and heat capacity of the calorimeter is 125 J/oC)
You have 50.0 g of warm water in a calorimeter at 35.0 oC. You then add 50.0 g of cold water at 1.7 oC and the final temperature of the system was 18.9 oC. (Specific Heat of Water = 4.184 J/(goC)) a.Calculate the q for warm water in kJ (should be a negative number) b.Calculate the q for cold water in kJ (should be a positve number) c.Using the answers for a. and b., calculate q for the calorimeter (should...
1. How much energy is need to change 125 grams of water from 25 oC to 98 oC ? 2. How much energy is given off when 200 grams of ethanol are cooled from 65 oC to 13 oC? Specific Heat of ethanol is 2.46 J/g oC. 3. The heat capacity of iron is 0.444 J/goC. What is the final temperature when 20 grams of iron wire at 80 oC losses 577.2 J? 4. A 56.0 grams of copper cylinder...
Final Temperature is 125 g
Use the table below to identify the type of metal in the object WEC Wwter 134 10 Copper ONS 0.0920 0.214 0.128 0.306 00125 Unknown metall 29.58 180°C 125 g 20.00°C Mass of metal (8) Initial temperature of metal ["C) Mass of water in the calorimeter (e) Initial temperature of water in the calorimeter (°C) Final temperature of water and metal ("C) Change in water temperature (AT . C) Energy absorbed by water (www.) Energy...
Initial Final Mass of Marshmallow (g) 7.1g 6.3g Temperature (oC) 29.1 oC 32.2 oC 1. Calculate the number of Joules released per gram of marshmallow burned. Show all work. Do this as follows: Find the T of water = Tfinal- Tinitial T of water = 3.1oC b. Find the Joules absorbed by the water using the formula below. Show all work. Joules = (10mL water) x (1.00g)/ mL x (3.1oC) x (4.184 joules)/goC The density of water is 1.00 g/mL, and...