January 11, 2010

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# January 11, 2010 - PowerPoint PPT Presentation

January 11, 2010. Learning Goal Clarify HW from this weekend Understand the concept of specific heat capacity In-Class Activities Semester 1 Review Quiz Pre-Lab Background Moodle Clarification HW Anything that did not get done this weekend! Specific Heat Capacity Pre-Lab

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Presentation Transcript
January 11, 2010
• Learning Goal
• Clarify HW from this weekend
• Understand the concept of specific heat capacity
• In-Class Activities
• Semester 1 Review Quiz
• Pre-Lab Background
• Moodle Clarification
• HW
• Anything that did not get done this weekend!
• Specific Heat Capacity Pre-Lab
• Weekly BLOG Post

### Specific Heat Capacity

Properties of Water

Background

On a sunny day, the water in a swimming pool might only warm up a degree or two while the concrete around the pool may become too hot to walk on with your bare feet. This may seem strange because both the concrete and the water are being heated by the same source – The Sun. This evidence suggests it takes more heat to raise the temperature of some substances than others. This, in fact, is true: The amount of heat needed to raise the temperature of 1 gram of a substance by 1°C is called the specific heat capacity, or simply the specific heat, of that substance. Water has a specific heat value that is higher in comparison with the specific heats for other materials, such as concrete. In this lab, the specific heat of a metal will be used to calculate the specific heat of water.

Definition

The amount of heat needed to raise the temperature of 1 gram of a substance by 1°C is called the specific heat capacity, or simply the specific heat, of that substance.

The formula

Q = cpmΔT

• Q (J) = heat, which is energy transferred into a heated substance (do not confuse this with temperature)
• cp (J/g ⁰C)= specific heat of a substance
• m (g)= mass of the heated substance
• ΔT (⁰C )= change in temperature of the heated substance
Practice problems

#1) How much heat is absorbed when 50.0 grams of a substance cools from 25.0 ⁰C to 12.0 ⁰C? The specific heat of the substance is 1.2 J/g ⁰C.

Using the formula: Q = cpmΔT

cp = 1.2 J/g ⁰C

m = 50.0 grams

ΔT = 25.0 ⁰C - 12.0 ⁰C= 13.0 ⁰C

Q= (1.2 J/g ⁰C)(50.0 g)(13.0 ⁰C) = 780 Joules

#2) What is the specific heat (cp ) of a 25 gram object that absorbs 500 Joules of energy, which causes the temperature of the object to change from 12⁰C to 22⁰C?

Rearranging the formula: cp = Q/(mΔT)

Q = 500 J

ΔT = 22⁰C - 12⁰C = 10⁰C

m= 25 g

cp = 500 J /(25g)(10oC) = 2 J/g ⁰C