Question

Water of mass, m at 100 ℃ is added to 0.50 kg of water at 20 ℃ in a lagged calorimeter of thermal capacity 105 JK −1 . If the specific heat capacity of water is 4200 kg −1K −1 and the final temperature of the mixture is 70℃, determine the value of m.

Answer 1

The value of m given that the temperature of the mixture is 70°C is : 0.875 Kg

Determine the value of m

Applying the principle of energy conservation

Heat lost by the hot body = heat absorbed by the cold water + calorimeter

= m * 4200 * ( 100 - 70  ) = 0.50 * 4200 * ( 70 -20 ) + 105 * ( 70 -20 )

= m * 126000 = 110250

therefore ;

m = 110250 / 126000

  = 0.875 kg

Hence we can conclude that The value of m given that the temperature of the mixture is 70°C is : 0.875 Kg

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