<p> Specific Heat Practice</p><p>How much energy does 250 mL of water gain when the temperature increases from 21.5 C to 33.7 C?</p><p>What is the change in energy when a 400 mL sample of water goes from 88.2 C to 68.7 C?</p><p>What is the specific heat of a 70 g piece of metal which looses 60 C while heating a 150 mL water sample up by 15 C?</p><p>What is the specific heat of a 200 g piece of metal heated in boiling water that heats a 250 mL sample of water from 19.5 C to 24.8 C?</p><p>What is the specific heat of a 100 g piece of metal cooled in the freezer to –20 C that cools a 500 mL sample of water from 19.8 C to 17.6 C?</p><p>EXTRA CREDIT</p><p>What will the new temperature be if a 300 g sample of metal with a specific heat of .8 J/1g 1C goes from boiling water into 100 g of water starting at 20 C? How much energy does 250 mL of water gain when the temperature increases from 21.5 C to 33.7 C? (4.18J/1g 1C)(250 mL water)(1 g water/1mL water)(12.2C) 12749 J</p><p>What is the change in energy when a 400 mL sample of water goes from 88.2 C to 68.7 C? (4.18J/1g 1C)(400 mL water)(1 g water/1mL water)(-19.5 C) -32604 J</p><p>What is the specific heat of a 70 g piece of metal which looses 60 C while heating a 150 mL water sample up by 15 C? Energy water gained = (4.18J/1g 1C)(150 mL water)(1 g water/1mL water)(15 C) 9405 J Therefore block lost 9405 J Cp = -9405 J/(70g)(-60 C) Cp =2.23 J/g C</p><p>What is the specific heat of a 200 g piece of metal heated in boiling water that heats a 250 mL sample of water from 19.5 C to 24.8 C? Energy water gained = (4.18J/1g 1C)(250 mL water)(1 g water/1mL water)(5.3 C) 5538.5 J Therefore block lost 5538.5 J Cp = -5538.5 J/(200 g)(-75.2 C) Cp =.37 J/g C</p><p>What is the specific heat of a 100 g piece of metal cooled in the freezer to –20 C that cools a 500 mL sample of water from 19.8 C to 17.6 C? Energy water lost = (4.18J/1g 1C)(500 mL water)(1 g water/1mL water)(-2.2 C) -4598 J Therefore block gained 4598 J Cp = 4598 J/(100 g)(37.6 C) Cp =1.22J/g C</p><p>EXTRA CREDIT</p><p>What will the new temperature be if a 300 g sample of metal with a specific heat of .8 J/1g 1C goes from boiling water into 100 g of water starting at 20 C? Energy water gains = -energy metal loses (4.18J/1g 1C)(100 mL water)(1 g water/1mL water)(t - 20) = -(.8J/1g 1C)(300 g)(t - 100) (418 J * t) - 8360 C= 24000 J C - (240J * t) (658 J * t)= 32360 J C t = 32360 J C/658 J t =49.18 C Ending temperature will be 49.18 C</p>