Presentation is loading. Please wait.

Presentation is loading. Please wait.

Determine the specific heat of the metal.

Published byElfreda Lamb Modified over 7 years ago

Similar presentations

Presentation on theme: "Determine the specific heat of the metal."— Presentation transcript:

1 Determine the specific heat of the metal.
If 85.0 g of a metal at 90.0 oC is added to g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•oC.) Determine the specific heat of the metal. Remember: calories gained by water = calories lost by metal

2 Determine the specific heat of the metal.
If 85.0 g of a metal at 90.0 oC is added to g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•K.) Determine the specific heat of the metal. calories gained by water = calories lost by metal 100.0 g 5.9 K H2O: 590 = cal Can we use  oC for  K?

3 Determine the specific heat of the metal.
If 85.0 g of a metal at 90.0 oC is added to g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•K.) Determine the specific heat of the metal. calories gained by water = calories lost by metal 100.0 g 5.9 K H2O: 590 = cal Now for the metal.

4 Determine the specific heat of the metal.
If 85.0 g of a metal at 90.0 oC is added to g of water at 25.0 oC, the final temperature of the mixture is 30.9 oC. (The specific heat of water is 1.00 cal/g•K.) Determine the specific heat of the metal. calories gained by water = calories lost by metal 100.0 g 5.9 K H2O: 590 = cal Sp.Ht. Metal 590 cal 0.117 85.0 g 59.1 oC

5 85. 0 g of a metal at 90. 0 oC is added to 100. 0 g of water at 25
85.0 g of a metal at 90.0 oC is added to g of water at 25.0 oC. Determine the final temperature of the mixture. (The specific heat of water is 1.00 cal/g•K and the specific heat of the metal is cal/g•K. ) calories gained by water = calories lost by metal 100.0 g (TF – TI) K 85.0 g (TI – TF)K = T must be positive

6 Insert the TI Values and Solve for TF
85.0 g of a metal at 90.0 oC is added to g of water at 25.0 oC. Determine the final temperature of the mixture. (The specific heat of water is 1.00 cal/g•K and the specific heat of the metal is cal/g•K. ) calories gained by water = calories lost by metal 100.0 g (TF – TI) K 85.0 g (TI – TF)K = T must be positive Insert the TI Values and Solve for TF TF = 30.0 oC

7

Download ppt "Determine the specific heat of the metal."

Similar presentations

Heat naturally flows from high temperature to low temperature.

Heat naturally flows from high temperature to low temperature.
Calculations involving Heat Themes: Conservation of Energy Thermal Equilibrium.

Calculations involving Heat Themes: Conservation of Energy Thermal Equilibrium.
How many Joules of energy are required to raise the temperature of 75 g of water from 20.0 o C to 70.0 o C? Heat =75 g x 50.0 o C x 4.18 J/g o C Heat=

How many Joules of energy are required to raise the temperature of 75 g of water from 20.0 o C to 70.0 o C? Heat =75 g x 50.0 o C x 4.18 J/g o C Heat=
EQ: Describe the parts and each part’s function in a calorimeter?

EQ: Describe the parts and each part’s function in a calorimeter?
Specific Heat Review Cp = q/(m)(Δt).

Specific Heat Review Cp = q/(m)(Δt).
Chapter 12.1. 1. Everything is made of particles 2. These particles move 3. Hot things move faster than cold things.

Chapter Everything is made of particles 2. These particles move 3. Hot things move faster than cold things.
Thermodynamics Ch 10 Energy Sections 10.4-10.6. Thermodynamics The 1st Law of Thermodynamics The Law of Conservation of Energy is also known as The 1st.

Thermodynamics Ch 10 Energy Sections Thermodynamics The 1st Law of Thermodynamics The Law of Conservation of Energy is also known as The 1st.
Heat And Specific Heat. Heat Energy that is transferred from one body to another because of temperature Unit 1 calorie (cal) – heat needed to raise 1g.

Heat And Specific Heat. Heat Energy that is transferred from one body to another because of temperature Unit 1 calorie (cal) – heat needed to raise 1g.
Temperature(“Degree of hotness”) - is the distribution of energy through the molecules. - related to the motion (speed) of molecules. Heat ≠ Temperature.

Temperature(“Degree of hotness”) - is the distribution of energy through the molecules. - related to the motion (speed) of molecules. Heat ≠ Temperature.
Ch 5: Thermochemistry The relationship between chemical reactions and energy changes involving heat is called thermochemistry.

Ch 5: Thermochemistry The relationship between chemical reactions and energy changes involving heat is called thermochemistry.
Heat calculations ∆H ~ ∆t ∆H ~ m ∆H=cm∆t +q Heat in -q Heat out ∆t.

Heat calculations ∆H ~ ∆t ∆H ~ m ∆H=cm∆t +q Heat in -q Heat out ∆t.
Reaction Energy and Reaction Kinetics Chapter 17 Notes.

Reaction Energy and Reaction Kinetics Chapter 17 Notes.
Conservation of energy PE = 1 j PE = 0 j (arbitrary) 1.0 m Work (w) = 1.O N ( 0.1 kg) X 1.0m = 1.0 j Arbitrary: assign KE = 0j Conservation of energy:

Conservation of energy PE = 1 j PE = 0 j (arbitrary) 1.0 m Work (w) = 1.O N ( 0.1 kg) X 1.0m = 1.0 j Arbitrary: assign KE = 0j Conservation of energy:
Energy! Energy: the capacity to do work or supplying heat Energy is detected only by its effects Energy can be stored within molecules etc as chemical.

Energy! Energy: the capacity to do work or supplying heat Energy is detected only by its effects Energy can be stored within molecules etc as chemical.
Heat: Lesson 4 Heat vs. Temperature. What happens to the movement of molecules as they’re heated? /energy-forms-and-changeshttp://phet.colorado.edu/en/simulation.

Heat: Lesson 4 Heat vs. Temperature. What happens to the movement of molecules as they’re heated? /energy-forms-and-changeshttp://phet.colorado.edu/en/simulation.
Chapter 11 Notes II Calorimetry. What is calorimetry? Essentially, the science of measuring heat change. At its most simple, you use the specific heat.

Chapter 11 Notes II Calorimetry. What is calorimetry? Essentially, the science of measuring heat change. At its most simple, you use the specific heat.
General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 2.5 Specific Heat Chapter 2 Energy and Matter © 2013 Pearson Education, Inc.

General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 2.5 Specific Heat Chapter 2 Energy and Matter © 2013 Pearson Education, Inc.
Thermochemistry Specific Heat.

Thermochemistry Specific Heat.
Heat Transfer Problems. 11-7 – Case #1 Silver Spoon Steel Spoon Boiling Water Baths at 100 o C m = 10.0 g s = 0.233 J/ g o C s = 0.51 J/ g o C Ti = 25.0.

Heat Transfer Problems – Case #1 Silver Spoon Steel Spoon Boiling Water Baths at 100 o C m = 10.0 g s = J/ g o C s = 0.51 J/ g o C Ti = 25.0.
Thermochemical Calculations

Thermochemical Calculations

Similar presentations

Ads by Google