logging in or signing up Heat Transfer and Specific Heat shelley.deford Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 104 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 24, 2012 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Heat Transfer and Specific Heat: Goals for this lesson: Define heat and compare it to internal energy Restate the 2 nd law of thermodynamics Describe three methods of heat transfer Explain what specific heat capacity tells us about an object’s response to heat Solve problems involving specific heat capacity Heat Transfer and Specific HeatHeat: Heat Heat (Q) : Energy transferred between objects due to a difference in temperature Heat is measured in Joules , just like our other forms of energy. Note that when thermal energy is contained in objects, it’s called internal energy ; when it’s in the process of moving between objects, it’s called heat . The 2 nd law of thermodynamics states that energy tends to spread out – this is why heat always flows from hot objects to cold object.Three Methods of Heat Transfer: Three Methods of Heat Transfer Conduction : occurs between objects in direct contact with each other Examples: touching a hot stove, holding an ice cube Convection : occurs when a fluid (liquid or gas) circulates or moves Examples: hot oven, boiling rice-in-a-bag Radiation : occurs via electromagnetic (light) waves and does not require physical contact Examples: sunlight, heat lampsSome Food for Thought…: Some Food for Thought… You get into your car on a super hot day, grab the seatbelt to put it on, and take a swig from a water bottle that’s been sitting in the cupholder all afternoon. What do you notice? When you’re eating a slice of pizza, what’s most likely to burn you – the crust, the sauce, or the cheese?Specific Heat Capacity: Specific Heat Capacity Specific heat (c): the amount of energy needed to change the temperature of some mass of a substance by one degree Celsius. The lower the specific heat, the more responsive an object is to heat energy. The higher the specific heat, the more an object will “store” that heat energy, rather than using it to increase in temperature.Specific Heat Formula: Specific Heat Formula Q = heat energy transferred (J) m = mass of substance (kg) c = specific heat (J/kg C) T = change in temperature (C) Q = mc T T = T f - T iSample Problem #1: Sample Problem #1 How much heat energy is needed to raise the temperature of 1.8 kg of water from 10 C to 80 C? (water’s specific heat is 4180 J/kgC) m = 1.8 kg c = 4180 J/kgC T i = 10 C T f = 80 C ∆T = 70 C Q = ? Q = mc∆T Q = (1.8 kg)( 4180 J/kgC)(70 C) Q = 526680 JSample Problem #2: Sample Problem #2 A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC) iron water m = 0.12 kg m = ? c = 450 J/kgC c = 4180 J/kgC T i = 98.9 C T i = 25 C T f = 30.1 C T f = 30.1 C ∆T = -68.8 C ∆T = 6.1 C Q = ? Q = ?Sample Problem #2: Sample Problem #2 Q iron = m iron c iron ∆T iron Q iron = (0.12 kg)(450 J/kgC)(-68.8 C) Q iron = -3715.2 J Q water = -Q iron = 3715.2 J A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC)Sample Problem #2: Sample Problem #2 A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC) iron water m = 0.12 kg m = ? c = 450 J/kgC c = 4180 J/kgC T i = 98.9 C T i = 25 C T f = 30.1 C T f = 30.1 C ∆T = -68.8 C ∆T = 6.1 C Q = -3715.2 J Q = 3715.2 JSample Problem #2: Sample Problem #2 Q water = m water c water ∆T water 3715.2 J = (m water )(4180 J/kgC)(6.1C) m water = ____ 3715.2 J____ (4180 J/kgC)(6.1C) m water = 0.146 kg A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC) You do not have the permission to view this presentation. In order to view it, please contact the author of the presentation.
Heat Transfer and Specific Heat shelley.deford Download Post to : URL : Related Presentations : Share Add to Flag Embed Email Send to Blogs and Networks Add to Channel Uploaded from authorPOINT lite Insert YouTube videos in PowerPont slides with aS Desktop Copy embed code: (To copy code, click on the text box) Embed: URL: Thumbnail: WordPress Embed Customize Embed The presentation is successfully added In Your Favorites. Views: 104 Category: Education License: All Rights Reserved Like it (0) Dislike it (0) Added: January 24, 2012 This Presentation is Public Favorites: 0 Presentation Description No description available. Comments Posting comment... Premium member Presentation Transcript Heat Transfer and Specific Heat: Goals for this lesson: Define heat and compare it to internal energy Restate the 2 nd law of thermodynamics Describe three methods of heat transfer Explain what specific heat capacity tells us about an object’s response to heat Solve problems involving specific heat capacity Heat Transfer and Specific HeatHeat: Heat Heat (Q) : Energy transferred between objects due to a difference in temperature Heat is measured in Joules , just like our other forms of energy. Note that when thermal energy is contained in objects, it’s called internal energy ; when it’s in the process of moving between objects, it’s called heat . The 2 nd law of thermodynamics states that energy tends to spread out – this is why heat always flows from hot objects to cold object.Three Methods of Heat Transfer: Three Methods of Heat Transfer Conduction : occurs between objects in direct contact with each other Examples: touching a hot stove, holding an ice cube Convection : occurs when a fluid (liquid or gas) circulates or moves Examples: hot oven, boiling rice-in-a-bag Radiation : occurs via electromagnetic (light) waves and does not require physical contact Examples: sunlight, heat lampsSome Food for Thought…: Some Food for Thought… You get into your car on a super hot day, grab the seatbelt to put it on, and take a swig from a water bottle that’s been sitting in the cupholder all afternoon. What do you notice? When you’re eating a slice of pizza, what’s most likely to burn you – the crust, the sauce, or the cheese?Specific Heat Capacity: Specific Heat Capacity Specific heat (c): the amount of energy needed to change the temperature of some mass of a substance by one degree Celsius. The lower the specific heat, the more responsive an object is to heat energy. The higher the specific heat, the more an object will “store” that heat energy, rather than using it to increase in temperature.Specific Heat Formula: Specific Heat Formula Q = heat energy transferred (J) m = mass of substance (kg) c = specific heat (J/kg C) T = change in temperature (C) Q = mc T T = T f - T iSample Problem #1: Sample Problem #1 How much heat energy is needed to raise the temperature of 1.8 kg of water from 10 C to 80 C? (water’s specific heat is 4180 J/kgC) m = 1.8 kg c = 4180 J/kgC T i = 10 C T f = 80 C ∆T = 70 C Q = ? Q = mc∆T Q = (1.8 kg)( 4180 J/kgC)(70 C) Q = 526680 JSample Problem #2: Sample Problem #2 A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC) iron water m = 0.12 kg m = ? c = 450 J/kgC c = 4180 J/kgC T i = 98.9 C T i = 25 C T f = 30.1 C T f = 30.1 C ∆T = -68.8 C ∆T = 6.1 C Q = ? Q = ?Sample Problem #2: Sample Problem #2 Q iron = m iron c iron ∆T iron Q iron = (0.12 kg)(450 J/kgC)(-68.8 C) Q iron = -3715.2 J Q water = -Q iron = 3715.2 J A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC)Sample Problem #2: Sample Problem #2 A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC) iron water m = 0.12 kg m = ? c = 450 J/kgC c = 4180 J/kgC T i = 98.9 C T i = 25 C T f = 30.1 C T f = 30.1 C ∆T = -68.8 C ∆T = 6.1 C Q = -3715.2 J Q = 3715.2 JSample Problem #2: Sample Problem #2 Q water = m water c water ∆T water 3715.2 J = (m water )(4180 J/kgC)(6.1C) m water = ____ 3715.2 J____ (4180 J/kgC)(6.1C) m water = 0.146 kg A 0.12-kg iron sample at 98.9 C is placed into a container of water at 25 C. Both the water and the iron reach a final temperature of 30.1 C. What mass of water is inside the container? (c iron = 450 J/kgC; c water = 4180 J/kgC)