a What is the final temperature when the two become equal Assume that coffee

# A what is the final temperature when the two become

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(a) What is the final temperature when the two become equal? Assume that coffee has the same specific heat as water. (b) The first time a student solved this problem she got an answer of 88 °C. Explain why this is clearly an incorrect answer.Solution (a) because of the law of conservation of energy, we write: q spoon + q coffee = 0 q spoon = – q coffee c spoon × m spoon × Δ T = c coffee × m coffee × Δ T 0.88 J/g °C × 45 g × ( T f – 24 °C) = –4.184 J/g °C × 180 g × ( T f – 85 °C) 39.6 T f – 950.4 = –753.12 T f + 64,015 753.12 T f + 39.6 T f = 64,015 + 950.4 792.7 T f = 64,965.4 T f = 82 °C; (b) The temperature of the coffee cannot increase when a colder spoon is placed in it. 26. When 50.0 g of 0.200 M NaCl( aq ) at 24.1 C is added to 100.0 g of 0.100 M AgNO 3 ( aq ) at 24.1 C in a calorimeter, the temperature increases to 25.2 C as AgCl( s ) forms. Assuming the specific heat of the solution and products is 4.20 J/g C, calculate the approximate amount of heat in joules produced. Solution
q = cm Δ T = 4.20 J/g °C × 150 g × (25.2 – 24.1) °C = 693 J 30. When 1.0 g of fructose, C 6 H 12 O 6 ( s ), a sugar commonly found in fruits, is burned in oxygen in a bomb calorimeter, the temperature of the calorimeter increases by 1.58 C. If the heat capacity of the calorimeter and its contents is 9.90 kJ/ C, what is q for this combustion? Solution q comb + q cal = 0 q comb = – q cal = –9.90 kJ/°C (1.58 °C) = –15.6 kJ The negative sign indicates that this was reaction exothermic, as expected. The mass did not directly enter into the calculation because the heat capacity of the system is calculated on a kJ/°C basis. The mass of the calorimeter is constant. 34. A teaspoon of the carbohydrate sucrose (common sugar) contains 16 Calories (16 kcal). Whatis the mass of one teaspoon of sucrose if the average number of Calories for carbohydrates is 4.1 Calories/g? 36. A pint of premium ice cream can contain 1100 Calories. What mass of fat, in grams and pounds, must be produced in the body to store an extra 1.1 × 10 3 Calories if the average number of Calories for fat is 9.1 Calories/g? Solution The average heat of combustion of fats is –9.1 kcal/g = –9.1 Calories/g: 2 1.0 g Mass = 1100 Calories = 121 g 1.2 10 9.1 Calories 10 lb Mass 121 g 0.27 lb 453.6 g . 38. Which is the least expensive source of energy in kilojoules per dollar: a box of breakfast cereal that weighs 32 ounces and costs \$4.23, or a liter of isooctane (density, 0.6919 g/mL) that costs \$0.45? Compare the nutritional value of the cereal with the heat produced by combustion of the isooctane under standard conditions. A 1.0-ounce serving of the cereal provides 130 Calories. Solution Convert ounces to grams and liters to grams, and then find the cost per gram. Determine the heat of combustion of isooctane on a per gram basis, and compare it with the cost of the heat produced by cereal.

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• Summer '19