A 0.0255 kg silver ring (Cp= 234 J kg-1ºC-1) is heated to 84 ºC and then placed in a calorimeter containing 0.05 kg of water at 24 ºC. The calorimeter is not perfectly insulated, so 0.14 kJ of energy is transferred to the surroundings before a final temperature is reached. What is the final temperature?
Answers
Given : A 0.0255 kg silver ring (Cp= 234 J kg¯¹ºC¯¹) is heated to 84 ºC and then placed in acalorimeter containing 0.05 kg of water at 24 ºC. The calorimeter is not perfectly insulated, so 0.14 kJ of energy is transferred to the surroundings before a final temperature is reached.
To find : The final temperature.
solution : from calorimetry law of conservation of energy,
heat lost by silver ring = heat gained by water + heat transferred to surroundings.
⇒0.0255 kg × 234 J/kg/°C × (84 - T) = 0.05 kg × 4200 J/kg/°C × (T - 24) + 0.14 kJ
⇒5.967(84 - T) = 210(T - 24) + 140
⇒501.228 - 5.967T = 210T - 5040 + 140
⇒501.228 + 4900 = 215.967 T
⇒T = 5401.228/215.967 = 25 ° C
Therefore the final temperature is 25 ° C
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