Unit 7: Thermochemistry

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Presentation transcript:

Unit 7: Thermochemistry Board Races

Measuring Heat A food item contains 124 nutritional Calories. How many calories does the food item contain? 124 × 1000 = 124,000 calories

Measuring Heat Ethanol is being used as an additive to gasoline. The combustion of 1 mole of ethanol releases 1367 kJ of energy. How many Calories are released? 1367 KJ 1 Calorie = 327 Calories 4.184 kJ

Measuring Heat To vaporize 2.00 g of ammonia, 656 calories are required. How many kilojoules are required? 656 cal 4.184 J 1 kJ = 2.74 kJ 1 cal 1000 J

Measuring Heat A 25.0 g bolt made of an alloy absorbed 250 J of heat as its temperature changed from 25.0ºC to 78.0ºC. What is the specific heat of the alloy? (units are J/g•ºC) 78.0ºC – 25.0ºC = 53.0ºC 250 J =0.189 J/g•ºC 25.0 g • 53.0ºC

Measuring Heat How many joules of heat are lost by 3580 kg of granite as it cools from 41.2ºC to -12.9ºC? The specific heat of granite is 0.803 J/g•ºC. (Watch your units!) -12.9ºC –3580 × 1000 = 3,580,000 g 41.2ºC = -54.1ºC q = (3,580,000 g)(0.803 J/g•ºC)(54.1ºC) q = 1.56×108 J

Measuring Heat How much heat is absorbed by a 44.7 g piece of lead when its temperature increases by 65.4ºC? (specific heat of lead is 0.129 J/g•ºC) q = (44.7 g)(0.129 J/g•ºC)(65.4ºC) q = 377 J

Measuring Heat When 10.2 g of canola oil at 25ºC is places in a wok, 3.34 kJ of heat is required to heat is to a temperature of 196.4ºC. What is the specific heat of canola oil? 3.34 kJ × 1000 = 3340 J 196.4ºC - 25ºC = 171.4ºC C = 1.91 J/g•ºC C = q 3340 J g • ΔT 10.2 g•171.4ºC