Lesson_1.2_Homework_and_Solutions

Lesson_1.2_Homework_ - Lesson 1.2 Homework Solutions 1 v = 108 km.h-1 = 30 m.s-1 The work done to bring the car to rest is W =(1/2)mv2 = 600 302 =

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Lesson 1.2 Homework Solutions 1. v = 108 km.h -1 = 30 m.s -1 . The work done to bring the car to rest is: W = (1/2)mv 2 = 600 × 30 2 = 540 kJ. Assuming that all this work is converted to heat, the heat generated is: Q = 540 kJ . 2. The amount of heat supplied by the heater in 15 min is: Q = 1500 × 15 × 60 = 1.35 × 10 6 J. If m is the mass of water that can be heated from 20 o c to 60 o C using this amount of heat, we have: 1.35 × 10 6 = mc ∆θ = m × 4180 × 40 = 167200 m 6 1.35 10 167200 8.1 kg m × = = 3. Here copper block loses heat and the aluminum cup and water gain heat. Let θ be the equilibrium temperature. Heat lost by copper = m cu c cu ∆θ = 0.245 × 390 (300 – θ ) Heat gained by aluminum cup = m al c al ∆θ = 0.15 × 900( θ – 20) Heat gained by water = m w c w ∆θ = 0.82 × 4180( θ – 20) Total heat gained = total heat lost 0.245 × 390 (300 – θ ) = 0.15 × 900( θ - 20) + 0.82 × 4180( θ – 20) θ = 27.3 o C 4. Heat lost by the substance = 0.215 c (330 – 43) = 61.71c where c is
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This note was uploaded on 05/01/2011 for the course PHY 2049 taught by Professor George during the Spring '11 term at Edison State College.

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Lesson_1.2_Homework_ - Lesson 1.2 Homework Solutions 1 v = 108 km.h-1 = 30 m.s-1 The work done to bring the car to rest is W =(1/2)mv2 = 600 302 =

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