![13. In How Hot is the Water? investigation, you were asked to determine the initial temperature of a sample of water by using thermodynamic relationships. Use this method to solve the following question. The temperature of 500 g of hot water in a calorimeter decreases to 45.0 °C after adding 30.0 g of 20 °C cold water. What was the temperature of the hot water in the calorimeter before adding the cold water? [The specific heat of water is 4184 J/g·°C].](http://img.homeworklib.com/questions/b3c30ce0-6c13-11ec-849e-75f387453702.png?x-oss-process=image/resize,w_560)
Homework Answers
Add Answer to:
13. In "How Hot is the Water? investigation, you were asked to determine the initial temperature...
A student heats 84.17 mL of water to 95.27°C using a hot plate. The heated water...
A student heats 84.17 mL of water to 95.27°C using a hot plate. The heated water is added to a calorimeter containing 73.92 mL of cold water. The water temperature in the calorimeter rises from J 2.15°C to 40.70°C. The specific heat capacity of water is 4.184 and the density of water is g. °C 1.00 mL Assuming that heat was transferred from the hot water to the cold water and the calorimeter, determine the heat capacity of the calorimeter....
A student heats 84.17 mL of water to 95.27°C using a hot plate. The heated water...
A student heats 84.17 mL of water to 95.27°C using a hot plate. The heated water is added to a calorimeter containing 73.92 mL of cold water. The water temperature in the calorimeter rises from J 2.15°C to 39.32°C. The specific heat capacity of water is 4.184 and the density of water is g•°C 1.00 mL Assuming that heat was transferred from the hot water to the cold water and the calorimeter, determine the heat capacity of the calorimeter. Heat...
Useful equations and constants: heat lost/gained = (Volume (Density)(AT)(specific heat capacity) (heat lost by hot...
Useful equations and constants: heat lost/gained = (Volume (Density)(AT)(specific heat capacity) (heat lost by hot water) - Cheat gained by cold water) calorimeter constant = Density co = 1.00 g/mL Specific heat capacity (water) - 4.184 J/g °C 1. What is thermochemistry? (2 pts) 2. A student adds 5.00 mL of cold water to 5.00 mL of hot water in a test tube surrounded by a Styrofoam installation (ie, calorimeter). The initial temperature of the cold water was 15.26 °C,...
structure.com/courses/1253444/quizzes/2442703/take Calculate the amount of heat water absorbs from a piece of hot metal using the...
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...
show all work please! 1. A piece of hot lead is added to cold water in...
show all work please! 1. A piece of hot lead is added to cold water in a constant volume calorimeter . The lead loses 3450 J of heat and the water gains 2890 J of heat. Calculate the mass of lead added if the initial temperature of the lead (0.158 J/g- degrees C) is 98 degrees C and the final temperature is 37 degrees C 2. Calculate the heat capacity of the calorimeter if the initial and final temperature of...
A hot lump of 32.3g of Copper at an initial temperature of 96.5 degrees Celsius in...
A
hot lump of 32.3g of Copper at an initial temperature of 96.5
degrees Celsius in 50mL H2O initially at 25.0 degrees Celsius and
allowed to reach thermal equilibrium. What is the final temperature
of the copper and water, given that the specific hear is 0.385J/g°C
and the specific heat of water is 4.184J/g°C?
4. A hot lump of 41.3 g of copper at an initial temperature of 94.8 °C is placed in 50.0 mL H2O initially at 25.0 °C...
A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed...
A
hot lump of 32.3 g of copper at an initial temperature of 96.5°C is
placed in 50.0 mL H2O initially at 25.0°C and allowed to reach
thermal equilibrium. What is the final temperature of the copper
and water given that the specific heat of copper is 0.385J/g°C and
the specific heat of water is 4.184J/g°C?
4. A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed in 50.0 mL H20 initially at 25.0°C...
Calorimetry Lab -Finding speficic heat capacity of calorimeter C(cal) I need this by today pls help!...
Calorimetry Lab
-Finding speficic heat capacity of calorimeter C(cal)
I need this by today pls help!
Exp.5 CHEMICAL EQUILIBRI 2. (2 marks) Table 1: Determining the heat capacity of the calorimeter. Run 1 Run 2 Run 3 Mass of the hot water, m. (g): Initial temperature of the hot water, T. (°C): Mass of the cold water, me (g): Initial temperature of the cold water and calorimeter, T. (°C): Final temperature of the mixture, T. (°C): Heat lost by the...
A 25 g cube of metal was heated in a hot temperature bath to 80oC and then added to 50 mL of deionized water. Using the following graph, explain how you would determine the temperature change tha...
A 25 g cube of metal was heated in a hot temperature bath to 80oC and then added to 50 mL of deionized water. Using the following graph, explain how you would determine the temperature change that has occurred in the system. The metal was added to the calorimeter (Styrofoam cup with water) at 180 seconds. What is the initial temperature of the water and the metal? What is the final temperature of the metal and the water? What occurred...
Please help how to solve this question An unknown metal weighting 13.11 g at a temperature...
Please help how to solve this question An unknown metal weighting 13.11 g at a temperature of 81.0 degrees Celius is place in a calorimeter containing 25.0 mL of water at a temperature of 25.0 degrees Celsius. The final equilibrium temperature is 30.0 degrees Celius. What s the specific heat of the metal? The specific heat of water is 4.184 J/g-C. Assume the density of water is 1.00g/mL and that there is no heat lost or gained by the calorimeter.
A student heats 84.17 mL of water to 95.27°C using a hot plate. The heated water is added to a calorimeter containing 73.92 mL of cold water. The water temperature in the calorimeter rises from J 2.15°C to 40.70°C. The specific heat capacity of water is 4.184 and the density of water is g. °C 1.00 mL Assuming that heat was transferred from the hot water to the cold water and the calorimeter, determine the heat capacity of the calorimeter....
A student heats 84.17 mL of water to 95.27°C using a hot plate. The heated water is added to a calorimeter containing 73.92 mL of cold water. The water temperature in the calorimeter rises from J 2.15°C to 39.32°C. The specific heat capacity of water is 4.184 and the density of water is g•°C 1.00 mL Assuming that heat was transferred from the hot water to the cold water and the calorimeter, determine the heat capacity of the calorimeter. Heat...
Useful equations and constants: heat lost/gained = (Volume (Density)(AT)(specific heat capacity) (heat lost by hot water) - Cheat gained by cold water) calorimeter constant = Density co = 1.00 g/mL Specific heat capacity (water) - 4.184 J/g °C 1. What is thermochemistry? (2 pts) 2. A student adds 5.00 mL of cold water to 5.00 mL of hot water in a test tube surrounded by a Styrofoam installation (ie, calorimeter). The initial temperature of the cold water was 15.26 °C,...
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...
A
hot lump of 32.3g of Copper at an initial temperature of 96.5
degrees Celsius in 50mL H2O initially at 25.0 degrees Celsius and
allowed to reach thermal equilibrium. What is the final temperature
of the copper and water, given that the specific hear is 0.385J/g°C
and the specific heat of water is 4.184J/g°C?
4. A hot lump of 41.3 g of copper at an initial temperature of 94.8 °C is placed in 50.0 mL H2O initially at 25.0 °C...
A
hot lump of 32.3 g of copper at an initial temperature of 96.5°C is
placed in 50.0 mL H2O initially at 25.0°C and allowed to reach
thermal equilibrium. What is the final temperature of the copper
and water given that the specific heat of copper is 0.385J/g°C and
the specific heat of water is 4.184J/g°C?
4. A hot lump of 32.3 g of copper at an initial temperature of 96.5°C is placed in 50.0 mL H20 initially at 25.0°C...
Calorimetry Lab
-Finding speficic heat capacity of calorimeter C(cal)
I need this by today pls help!
Exp.5 CHEMICAL EQUILIBRI 2. (2 marks) Table 1: Determining the heat capacity of the calorimeter. Run 1 Run 2 Run 3 Mass of the hot water, m. (g): Initial temperature of the hot water, T. (°C): Mass of the cold water, me (g): Initial temperature of the cold water and calorimeter, T. (°C): Final temperature of the mixture, T. (°C): Heat lost by the...
Most questions answered within 3 hours.
-
Research new software products that assist in project
communication management. Write a paper summarizing your findings,...
asked 10 minutes ago -
the figure below to the left shows a beaker completely filled
with water. then a block...
asked 8 minutes ago -
A certain rifle bullet has a mass of 5.25 g. Calculate the de
Broglie wavelength of...
asked 12 minutes ago -
what is the duistribution channels of HUL in india and how they
distribute?
how is this...
asked 15 minutes ago -
Batteries play an important role in our everyday lives. From
making many devices mobile, to allowing...
asked 27 minutes ago -
It is possible to iterate through an array quickly and easily
using a loop?
asked 35 minutes ago -
Determine the pHpH of an HFHF solution of each of the following
concentrations. The KAKA value...
asked 31 minutes ago -
Two point charges of 7.0mL are initially 11.5m apart. What is
the change in potential energy...
asked 32 minutes ago -
Please write as a short answer
You have heartburn and a two hour working session on...
asked 36 minutes ago -
Please solve with excel and mention the excel function.
Thanks
Kevin deposits $10,000 in a savings...
asked 38 minutes ago -
Which Special Journal would the following transaction be
journalized in?
Sold merchandise on account to Customer...
asked 50 minutes ago -
Assuming 100% dissociation, calculate the freezing point (?fTf)
and boiling point (?bTb) of 1.45 ? K3PO4(aq)1.45...
asked 52 minutes ago