When a 290-g piece of iron at 170 ∘C is placed in a 95-gg aluminum calorimeter cup containing 250 g of liquid at 10∘C, the final temperature is observed to be 32 ∘C. The value of specific heat for iron is 450 J/kg⋅C∘, and for aluminum is 900 J/kg⋅C∘
Part A
Determine the specific heat of the liquid.
Express your answer using two significant figures.
heat lost by iron = heat gained by cup + liquid
miron c iron ( Ti - Tf ) = mAl cAl (Tf - Ti ) + m liquid c liquid (Tf - Ti )
0.290 * 450 * ( 170 - 32) = 0.095 * 900 * (32 - 10) + 0.250 * c liquid ( 32 - 10)
0.290 * 450 * 138 = 0.095 * 900 * 22 + 0.250 * c liquid * 22
(0.290 * 450 * 138) - (0.095 * 900 * 22) = 0.250 * c liquid * 22
16128 = 0.250 * c liquid * 22
c liquid = 2932.4 J/Kg.oC
or
c liquid = 2900 J/Kg.oC
or
c liquid = 2.9e3 J/Kg.oC
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