HW09-key - Chemistry 111 - Key! Homework #9 Due: November...

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Chemistry 111 - Key! Homework #9 Due: November 09, 2005 (Wednesday) 1. The first law of thermodynamics states that… a) the energy of every pure substance is zero. b) the disorder is always increasing. c) enthalpy is always increasing. d) the total energy of the universe is increasing. e) the entropy of the surroundings is zero. f) the total energy of the universe is constant. 2. How many calories are equivalent to 364 J? a) 1523 cal b) 87.0 cal c) 14.6 cal d) 0.364 cal e) 1.33 cal f) None of these. This is a simple conversion factor. 364 J × 1 cal 4.184 J = 87.0 cal . 3. Specific heat is… a) the number of kelvins that 1.00 mol of a substance is raised by heating it for 1.00 minute. b) the amount of heat needed to change 1.00 mol of a substance’s temperature by 1.00 K. c) the amount of energy required to melt 1.00 g of a substance. d) the amount of substance that is heated by 1.00 K. e) the number of kelvins that 1.00 g of a substance is raised by heating it for 1.00 minute. f) the amount of heat needed to change 1.00 g of a substance’s temperature by 1.00 K. 4. A system consists of 650 g of water originally at 20.5°C. How much heat must the system absorb to raise its temperature to 65.5°C? a) 151 kJ b) 55.8 kJ c) 178 kJ d) 122 kJ e) 131 kJ f) None of these.
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2 Homework #9 / Fall 2005 Answer Key This is a simple problem based on the simple fact that water is a pure substance and that the amount of heat that it will absorb in order to warm up depends on its specific heat. q = m c Δ T. = 650 g (4.184 J/g deg) (65.5°C – 20.5°C) = 122,000 J or 122 kJ . 5. Equal masses of two substances, A and B, each absorb 25 J of energy. If the temperature of A increases by 4 degrees while the temperature of B increases by 8 degrees, one can safely say that… a) the specific heat of B is double that of A. b) the specific heat of A is double that of B. c) the specific heat of B is negative. d) the specific heat of A is negative. e) the specific heat of B is triple that of A. 6. Consider a dropping mass of 500 kg which turns a paddle as it falls. Calculate the height that the mass should drop in order to increase the temperature of 650 g of water, originally at 20.5°C, to 65.5°C. W = - mass (g) ( Δ h); g = 9.81 m/s 2 a) 26,700 m b) 23.3 m c) 36.3 m d) 26.7 m e) 24.9 m f) None of these. We immediately recognize that the transformation in this question has the exact same initial and final conditions as in the previous problem. This means that both problems have exactly the same Δ E. The Δ E for the question above was 122 kJ, therefore, it takes the same amount of energy in order to make this transformation, although it will all be in the form of work. W
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HW09-key - Chemistry 111 - Key! Homework #9 Due: November...

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