HomeworkLib
Question

2. A volume of 34.7 mL of H2O is initially at 28.0 oC. A chilled glass...

2. A volume of 34.7 mL of H2O is initially at 28.0 oC. A chilled glass marble weighing 4.00 g with a heat capacity of 3.52 J/oC is placed in the water. If the final temperature of the system is 26  oC , what was the initial temperature of the marble? Water has a density of 1.00 g/mL and a specific heat of 4.18 J/goC.

Enter your answer numerically, to three significant figures and in terms of oC.

science   chemistry   
0 0
Add a comment Improve this question Transcribed image text
Next > < Previous
Sort answers by oldest

Homework Answers

Answer #1

m(marble) = 4.0 g

T(marble) = to be calculated

Since density of water is 1.00 g/mL and volume is 34.7 mL,

m(water) = 34.7 g

T(water) = 28.0 oC

C(water) = 4.18 J/goC

T = 26.0 oC

We will be using heat conservation equation

use:

heat gained by water = heat lost by marble

m(water)*C(water)*(T-T(water)) = m(marble)*C(marble)*(T(marble)-T)

34.7*4.18*(28.0-26.0) = 4.0*3.52*(26.0-T(marble))

T(marble)= 5.40 oC

Answer: 5.40 oC

0 0
Add a comment
Know the answer?
Add Answer to:
2. A volume of 34.7 mL of H2O is initially at 28.0 oC. A chilled glass...
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Log In

Sign Up

Not the answer you're looking for? Ask your own homework help question. Our experts will answer your question WITHIN MINUTES for Free.
Similar Homework Help Questions

  • A volume of 35.6 mL of H2O is initially at 28.0 oC. A chilled glass marble...

    A volume of 35.6 mL of H2O is initially at 28.0 oC. A chilled glass marble weighing 4.00 g with a heat capacity of 3.52 J/oC is placed in the water. If the final temperature of the system is 26.4  oC , what was the initial temperature of the marble? Water has a density of 1.00 g/mL and a specific heat of 4.18 J/goC.

  • A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled...

    A volume of 125 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.00  ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units.

  • Part A A volume of 115 mL of H2O is initially at room temperature (22.00 °C). A chilled steel rod at 2.00 °C is pl...

    Part A A volume of 115 mL of H2O is initially at room temperature (22.00 °C). A chilled steel rod at 2.00 °C is placed in the water. If the final temperature of the system is 21.10 °C, what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g. °C) specific heat of steel = 0.452 J/(g. °C) Express your answer to three significant figures and include the appropriate units. View Available...

  • PART A A volume of 110. mL of H2O is initially at room temperature (22.00 ∘C)....

    PART A A volume of 110. mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.40 ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units. PART B The...

  • Part a. A volume of 90.0 mL of H2O is initially at room temperature (22.00 ∘C)....

    Part a. A volume of 90.0 mL of H2O is initially at room temperature (22.00 ∘C). A chilled steel rod at 2.00 ∘C is placed in the water. If the final temperature of the system is 21.10  ∘C , what is the mass of the steel bar? Use the following values: specific heat of water = 4.18 J/(g⋅∘C) specific heat of steel = 0.452 J/(g⋅∘C) Express your answer to three significant figures and include the appropriate units. Part b. The specific...

  • QUESTION 17 3.00 g Na2CO3 is dissolved in H2O and titrated with HCl. 28.0 mL of...

    QUESTION 17 3.00 g Na2CO3 is dissolved in H2O and titrated with HCl. 28.0 mL of a HCl  solution were required to titrate the Na2CO3 solution. What is molarity of HCl solution? Na2CO3(aq) + 2HCl(aq) --> 2NaCl(aq) +   H2O(l) + CO2(g) 1.88 molar 3.21 molar 0.0417 molar 9.37 molar 2.59 molar 0.123 molar 0.553 molar 2.02 molar       8 points    QUESTION 18 A metal object at 28.0 oC is heated by gaining 2.31 kJ of heat from the environment. The final...

  • In the following experiment, a coffee-cup calorimeter containing 100 g of H2O is used. The initial...

    In the following experiment, a coffee-cup calorimeter containing 100 g of H2O is used. The initial temperature of the calorimeter is 23.0 oC. If 7.6 g of CaCl2is added to the calorimeter, what will be the final temperature of the solution in the calorimeter? Assume that the heat capacity of the solution is 4.18 J/goC, and that the heat capacity of the calorimeter is negligible. The heat of dissolution ΔHsoln of CaCl2 is −82.8 kJ/mol. Enter your answer numerically in...

  • 1) A calorimeter contains 28.0 mL of water at 11.5 ?C . When 2.20 g of...

    1) A calorimeter contains 28.0 mL of water at 11.5 ?C . When 2.20 g of X (a substance with a molar mass of 51.0 g/mol ) is added, it dissolves via the reaction X(s)+H2O(l)?X(aq) and the temperature of the solution increases to 30.0 ?C . Calculate the enthalpy change, ?H, for this reaction per mole of X. Assume that the specific heat of the resulting solution is equal to that of water [4.18 J/(g??C)], that density of water is...

  • Consider the dissolution of 1.50 grams of salt XY in 75.0 mL of water within a...

    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

  • A metal sample weighing 42.6 g and at a temperature of 100.0 oC was placed in...

    A metal sample weighing 42.6 g and at a temperature of 100.0 oC was placed in 40.6 g of water at 24.8oC. At equilibrium the final temperature of the water and metal was 35.0oC. a. What was the change in temperature for the water? oC b. What was the temperature change for the metal? oC c. Taking the specific heat of water to be 4.184J/goC,caculate the specific heat of the metal. J/goC d. What is the approximate molar mass of...

ADVERTISEMENT
Free Homework Help App
Download From Google Play
Scan Your Homework
to Get Instant Free Answers
Need Online Homework Help?
Ask a Question
Get Answers For Free
Most questions answered within 3 hours.
ADVERTISEMENT
ADVERTISEMENT