hw1_sol - 1 a. The heat capacity of liquid water is 1 cal/g...

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1 a. The heat capacity of liquid water is 1 cal/g K (from the definition of the calorie). m (g) × 1 (cal/g K) × (36.7 – 8)(K) = 200000 cal m = 6968.6g b. The ice should first be melted and then heated from T ice =0 0 C to T body =36.7 0 C 200000 cal = 1000 1 1 6.01( / ) 36.7 200000 1 4.184 18 / Jc a l x kJ mol K cal kJ J g mol  +=   , where x is in grams. x = 1716.7 g 2. a. W = 100 × mg h = 178 kJ b. W = 0 (mechanical work is equal to the change of the potential energy of the weight, which is zero because the weight returned to where it started) c. You surely know from your experience that method b is more efficient as a way of burning calories (you need extra calories to lower the weight) d. In case a U = (-W) + q (you use calories to do work and to produce heat). The work that was done by the barbell on you is equal to (–W), where W is the work you performed on the barbell. In the case b U = q, i.e. all of the burned calories are eventually dissipated as heat. According to what we have concluded in Part c,| U b | > | U a | 3. Using the equation of state,
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hw1_sol - 1 a. The heat capacity of liquid water is 1 cal/g...

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