Help please. General chemistry

label Chemistry
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A metal weighing 43.0 grams and having a specific heat of 0.315J/(g celsiuc) is heated to 92.8 Celsius. When this heated metal is transferred into a coffee cup calorimeter containing x grams of water at 12.5 Celsius the final temp of this combined system is found to be 18.3celsius. Determine the value of x. The specific heat of water =4.184J(g deg C)

Apr 10th, 2015

for metal

mc(t-tf)= 43*0.315(92.8-18.3)= 1009.1

for calorimeter

mc(tf-ti)=x*4.184*(18.3-12.5)=24.27x

heat is not lost

so 24.27x=1009.1

x=41.6 g........................................................best me

Apr 10th, 2015

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Apr 10th, 2015
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