Specific Heat The amount of heat energy needed to raise the temperature of a 1.0g of a substance 1oC.

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Presentation transcript:

Specific Heat The amount of heat energy needed to raise the temperature of a 1.0g of a substance 1oC.

Substances with low specific heats absorb & release heat E quickly Ex. Metals Aluminum 0.9 J/goC Those with high specific heats absorb & release heat E slowly Ex. Water 4.18 J/goC

Variables & their units . . . q is Heat energy absorbed or released measured in Joules Specific Heat (C) units: J/goC Heat of fusion (Hf ) E needed to melt one gram of a substance at constant temperature. Heat of vaporization (Hv) E needed to vaporize one gram of a substance at constant temperature.

This unknown substance has a mass of 50 This unknown substance has a mass of 50.0g and heat is added at a rate of 380.0J/min What is the specific heat of the liquid? q = mCΔT q = m = C = ? ΔT =

To find q you must multiply the rate by time on the graph. This unknown substance has a mass of 50.0g and heat is added at a rate of 380.0J/min What is the specific heat of the liquid? Let’s look at q. You know that q should be in Joules, but notice that a RATE is given in J/min. To find q you must multiply the rate by time on the graph. q = 380.0J/min ×12min = 4650J 12 min in the liquid phase

You need to read the graph and determine the temperature change This unknown substance has a mass of 50.0g and heat is added at a rate of 380.0J/min What is the specific heat of the liquid? Now let’s figure out ΔT. You need to read the graph and determine the temperature change WHILE THE SUBSTANCE IS IN THE LIQUID STATE. 40oC

2.28J/goC q = mCΔT m = C = ? ΔT = 4560J 50g 40oC q = This unknown substance has a mass of 50.0g and heat is added at a rate of 380.0J/min What is the specific heat of the liquid? The mass is given in the problem, so now you can substitute and solve. q = mCΔT q = m = C = ? ΔT = 4560J 50g 40oC 2.28J/goC

This unknown substance has a mass of 15g and heat is added at a rate of 540J/min Calculate the heat of fusion. q = mHf 9 min Heat of fusion is calculated where the substance is MELTING Because you are given the RATE, you must determine the TIME for melting. Q=(9 min)(540J/min)= 4860J

This unknown substance has a mass of 15g and heat is added at a rate of 540J/min Calculate the heat of fusion. Now substitute and solve. q = mHf 324 J/g 4860 J 15g ? q = m = Hf =

This unknown substance has a mass of 10g and heat is added at a rate of 0.35KJ/min Calculate the Heat of Vaporization. q = mHv Heat of VAPORIZATION is calculated from the part of the graph where the substance is VAPORIZING.

This unknown substance has a mass of 10g and heat is added at a rate of 0.35KJ/min Calculate the Heat of Vaporization. q = mHv 6 min VAPORIZiNG 2100 J 10g ? q = m = Hv = 210.0J/g

As you eat your pizza, answer the following questions. Mmmmm I want to make a pizza and I pre-heat the oven to 475oF (~246oC). As I get ready to put the pizza in the oven, I realize that I’ve left an Aluminum cookie sheet in there. Oops! Earlier I was washing dishes, and I left the sink full of water, now its gross and cold. I put ovenmits on and take out the hot cookie sheet, and put it into the water. Then put the pizza in the oven to cook. Finally I go to take care of the cookie sheet, and realize that the water is really really warm, and I don’t need ovenmits to touch the aluminum pan. Hmmmmm what happened? As you eat your pizza, answer the following questions.

How hot was the Aluminum pan in the oven? What was the approximate temperature of the “cold” water in the sink? What happened to the temps of the pan and the water when they came into contact with each other? Is this change the same? Where did the Energy come from that raised the temperature of the water?

The same temp as the oven 475oF (~246oC). How hot was the Aluminum pan in the oven? The same temp as the oven 475oF (~246oC). What was the approximate temperature of the “cold” water in the sink? If it sat for a while, probably 20oC room temp. What happened to the temps of the pan and the water when they came into contact with each other? The temp of the pan went down and the temp of the water went up. Is this change the same? NO! The temp of the water may have gone up to 24oC (ΔT of water = 4oC) while the temp of the pan came down to 24oC. (ΔT of pan = 222oC) Where did the Energy come from that raised the temperature of the water? The Energy came from the pan!

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