Suppose that one adds a piece of aluminum weighing 100 g. andheated to a tempera

Question

Suppose that one adds a piece of aluminum weighing 100 g. andheated to a temperature of 180 degree C. to a water calcrimetercontaining 200g. of water at 10 degreeC and with a water equivalentof 152 cal/deg. What will be the final temperature of thewell-stirred mixture? Specific heat of Aluminum is 0.22 cal/g-C Suppose that one adds a piece of aluminum weighing 100 g. andheated to a temperature of 180 degree C. to a water calcrimetercontaining 200g. of water at 10 degreeC and with a water equivalentof 152 cal/deg. What will be the final temperature of thewell-stirred mixture? Specific heat of Aluminum is 0.22 cal/g-C

Explanation / Answer

Heat Loss = Heat Gain 100 * .22 * T1 = (200 + 152)T2 T1 / T2 = 352 / 22 =16 ( 100 - Tf) = 16 ( Tf -10)        17 Tf =260      Tf = 15.3 deg T1 / T2 = 352 / 22 =16 ( 100 - Tf) = 16 ( Tf -10)        17 Tf =260      Tf = 15.3 deg

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