A lead ball of 250g to 210 degrees Celsius is placed in an aluminum calorimeter

Question

A lead ball of 250g to 210 degrees Celsius is placed in an aluminum calorimeter of 90g which contains 300g of liquid at 20 degrees Celsius. If the final temperature is 30 degrees C, what is the specific heat of the liquid? Answer: 1,68kJ/(kg•K) A lead ball of 250g to 210 degrees Celsius is placed in an aluminum calorimeter of 90g which contains 300g of liquid at 20 degrees Celsius. If the final temperature is 30 degrees C, what is the specific heat of the liquid? Answer: 1,68kJ/(kg•K) Answer: 1,68kJ/(kg•K)

Explanation / Answer

heat lost = heat gained

250*(210-30)*0.13 = 90*(30-20)*0.90+300*(30-20)*s

[250*(210-30)*0.13 - 90*(30-20)*0.90]/3000 =s

s = [250*(210-30)*0.13 - 90*(30-20)*0.90]/3000

   = 1.68 J/gK

   = 1.68 KJ/kg.K

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