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Name________________________________ Date_____________ Chp. 10.3 (part A) Notes 1. Specific Heat beakers at the same temperature are a. Some things ____heat__ up or supplied with the same amount of ___cool_____ down faster than others. energy, which will have the higher final b. Specific heat is the __amount_ of temperature? Ethyl Alcohol heat required to raise the temperature of 1 kg of a material by one degree (C or 3. A calorimeter is used to help K). measure the specific heat of a substance. 4. Specific heat capacity equation: The formula is __Q = mc∆T. Cp = specific heat capacity at constant pressure Q = energy transferred as heat m= mass ∆T= change in temperature 5. Place the variable in the triangle Specific How long below and write the equation for each Heat it takes to heat up variable. 664 J/kg˚C fast Q Land m x c x ∆T 4186 J/ kg˚C slow Water 2. Write a summary statement based on the information in the chart about how specific heat value determines how long a substance takes to heat up or cool down. The lower the specific heat value, the quicker it heats up. The A) The temperature change in ºC is higher the specific heat value the equal in magnitude to a 1K, so _NO____ slowe it heats up. conversion is necessary. Concept Question: Ethyl alcohol has about one-half the specific heat capacity of water. If equal masses of alcohol and water in separate Examples 4. Finding the final temperature with a) How much energy must be two substances transferred as heat to 420 kg of water in Energy absorbed by water = energy a bathtub in order to raise the water’s released by the substance temperature from 25°C to 37ºC? Qw = -Qx Specific Heat of Water = 4186 J/kg•ºC C1m1∆T1 = C2m2∆T2 m=420 Q=mc∆T ∆T=37-25=12 Q=420(12)4186 ∆T should Tf= c1m1ti1 + c2m2ti2 c=4186 Q = 2.11 x 107J always c1m1 + c2m2 be positive. b) What mass of water is required to absorb 4.7 x 105 J of energy from a car Teacher Example: engine while the temperature increases 1. What is the final temperature when a from 298K to 355K? Specific Heat of 3.0kg gold bar at 99˚C is dropped into Water = 4186 J/kg•ºC 0.22kg of water at 25˚C? Q=4.7 x 105 m=Q/(c∆T) m1=3 Tf= c1m1ti1 + c2m2ti2 ∆T=355-298=57 m=4.7 x 105 T1=99 c=4186 (4186 x 57) m2=0.22 c1m1 + c2m2 m=1.97kg T2=25 = (129x3x99 + 4186x0.22x25) Student Practice c1=129 (129x3 + 4186x.22) a) A vanadium bolt gives up 1124J of c2=4186 = 46.9˚C energy as its temperature drops 25K. If the bolt’s mass is 0.093kg, what is its 2. A 0.225kg sample of tin initially at specific heat? 97.5ºC is dropped into 0.115kg of water Given Formula Substitution Answer initially at 10.0ºC. If the specific heat Q=1124 c=Q/(∆Tm) capacity of tin is 230 J/kgºC, what is the ∆T=25 c=1124 =483J/kg˚C final temperature of the tin-water m=0.093 (25 x 0.093) mixture? m1=0.225 Tf= c1m1ti1 + c2m2ti2 b) How much energy must be added to a T1=97.5 bowl of 125 popcorn kernels in order for m2=0.115 c1m1 + c2m2 them to reach a popping temperature of T2=10 175ºC? Assume that their initial c1=230 temperature is 21ºC, that the specific c2=4186 heat capacity of popcorn is 1650J/kgºC, =(230x0.225x97.5+4186x0.115x10) and that each kernel has a mass of (230x0.225 + 4186x.115) 0.105g. = 18.5˚C Given Formula Substitution Answer ∆T=175-21=154 Q=mc˚∆T c=1650 =0.013(1650)154 m=0.013 = 3303J To determine mass you need to first convert 0.105g to kg by dividing by a 100. You will get 0.000105kg. Then you multiply that 0.000105kg by 125 to determine the total mass. Student Practice 3. What is the final temperature when 0.032kg of milk at 11ºC is added to 0.16kg of coffee at 91ºC? Assume the specific heat capacities of the two liquids are the same as water, and disregard any energy transfer to the liquids’ surroundings. Given Formula Substitution Answer m1=0.032 Tf= c1m1ti1 + c2m2ti2 T1=11 m2=0.16 c1m1 + c2m2 T2=91 c1=4186 c2=4186 =(4186x0.032x11+4186x0.16x91) (4186x0.032 + 4186x.16) = 77.7˚C

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posted: | 9/18/2012 |

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