advertisement

Entrance Ticket Part I 3/10/15 Add the following to 17.1 #5-9 1.) How much energy in Joules is needed to heat up 15.0 grams of water from 7.25 ºC to 38.0 ºC? (specific heat of water is 4.184 J/gºC) Entrance Ticket Part II 3/10/15 2.) When a certain metal is cooled from 130 ºC to 22 ºC it releases 415 Joules. What is the mass of the metal if its specific heat capacity is 0.24 J/gºC. 3.) How much energy is needed to melt a 100. g chunk of ice at 0 ºC? (heat of fusion is 334 J/g) Answers 2.) 3.) Q = mcΔT 415 J = m x 0.24 J/gºC x 108 ºC m = 16 g Q = m ΔHfus Q = 100. g x 334 J/g Q = 33,400 J Entrance Ticket 3/12/15 1.) 12,000 Joules of energy are needed to heat up a certain mass of water by 10. ºC. What is the mass of water? (specific heat of water is 4.18 J/gºC) Q = mcΔT 12,000 J = m x 4.18 J/gºC x 10. ºC m = 290 g Specific Heat Practice 1.) 24,000 J or 5,700 cal 2.) 8.9 x 106 J or 8,900,000 J 3.) 97 ºC 4.) 36.2 g 5.) 1,200 cal Specific Heat Practice 6.) 0.838 J/gºC 7.) 1,690 g 8.) 2,980 kJ 9.) 3.0 ºC 10.) 0.89 J/gºC Aluminum Heat Problems Cont… How many Joules would it take to heat 250. g of ice from 0ºC to 100 ºC? (specific heat of water is 4.184 J/g ºC & the heat of fusion is 334 J/g) Phase Δ & Temp. Δ Q = m ΔHfus Q = 250g x 334 J/g Q = 83500 J Q = mcΔT Q = 250g x 4.184 J/gºC x 100ºC Q = 104600 J Q = 83500 + 104600 = 188,100 J Heat Problems Cont…Honors How many Joules would it take to heat 125 g of ice from -10ºC to 95ºC? -specific heat of ice is 2.1 J/gºC, -specific heat of water is 4.184 J/gºC -the heat of fusion of H20 is 334 J/g -the freezing point of water is 0ºC Answer Q=125g x 2.1 J/gºC x 10ºC = 2,625 J Q= 125g x 334 J/g = 41,750 J Q=125g x 4.184 J/gºC x 95ºC = 49,685 J Total Q = 94,000 J -specific heat of ice is 2.1 J/gºC, -specific heat of water is 4.184 J/gºC -the heat of fusion of H20 is 334 J/g -the freezing point of water is 0ºC Heat Problems Cont…Honors How many Joules would it take to melt 65 g of ice that has a temp. of -7.5ºC? -specific heat of ice is 2.1 J/gºC, -the heat of fusion of H20 is 334 J/g -the freezing point of water is 0ºC Q = mcΔT Q = 65 g x 2.1 J/gºC x 7.5 ºC Q =1024 J Q = mΔHfus Q = 65 g x 334 J/g Q = 21,710 J 1024 J + 21,710 J = 23,000 J 17 Quiz Friday -Know what heat is. -Know how heat flows, and how it is related to kinetics. -Know how to solve phase change & temperature change problems involving heat transfer. -Know what endothermic and exothermic mean. -Know what latent heat is, and when it’s used. -Know what specific heat capacity is, and when it’s used. -1 Cal = 1000 cal -Understand your labs 150 Cheeto Lab p. 33 www.fritolay.com Cheeto Lab Objective: Determine the number of Calories per gram in a Cheeto. Materials: Soda can calorimeter (measures heat) Cheeto Pins Graduated cylinder Mass scale Water Thermometer Matches Cheeto Lab Procedure: Write your own procedure! Hints: Q = mcΔT cwater = 1 cal/g°C 1000 cal = 1 Calorie Assumption: The energy gained by the water will be equal to the energy lost by the Cheeto! Analysis: 1. Show your data and calculations. a. Calculate the heat transferred (the calories of your Cheeto) b. Convert to food Calories c. Divide by the mass of the Cheeto d. Compare to the food label 2. Calculate the % yield. (Standard 3f *) actual. yield theoretical. yield 100 percent. yield Cheeto Lab Grading: Every lab partner must have their own identical lab. Turn in one lab per group. All the labs must be complete before the one sample is turned in! Failure to create your copy of the lab will result in a zero on your lab grade. Cheeto Lab Grading: Procedure Analysis Safety Total 5 points 15 points (3 each) 5 points 25 points © 2007 BPHS Mr. Reule / Mr. Melton