398 CHAPTER 11 Energy in Thermal Processes 31. A 40-g block of ice is cooled to

Question

398 CHAPTER 11 Energy in Thermal Processes 31. A 40-g block of ice is cooled to C and is then added to 560 g of water in an 80-g copper calorimeter at a tem perature of 25°C. Determine the final temperature of the system consisting of the ice, water, and Calorimeter (If not all the ice melts, determine how much ice is left.) Remember that the ice must first warm to 00C, melt, and then continue warming as water. (The specific heat of ice is 0.500 cal/g 2 090 J/kg BO When you jog, most of the food energy you burn above your basal metabolic rate (BMR) ends up as

Explanation / Answer

Heat gained by ice= heat lost by calorimeter. Let final temperature be 0 C, and ice melts be x kg

0.040*2090*78 + 334000*x = 385*25*0.080 +4186*25*0.560 = 59374

x = [59374-0.040*2090*78]/334000 = 0.158 kg

which is more than mass of ice, so temperature is more than 0C

0.040*2090*[78+T] + 334000*0.040 = 385*[25-T]*0.080 +4186*0.560*[25-T]

  [78+T]*83.6+13360 = [25-T]* [385*0.080+4186*0.560]

T = 16.06 C answer and all ice melts

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