A 1.50 kg iron block heated to 140. 0 ∘ C is placed into 1.10 kg of water initially at 25. 0 ∘ C . After thermal equilibrium, the final temperature becomes 40. 0 ∘ C .Using this information, determine

Question

A 1.50 kg iron block heated to 140. 0 ∘ C is placed into 1.10 kg of water initially at 25. 0 ∘ C . After thermal equilibrium, the final temperature becomes 40. 0 ∘ C .Using this information, determine the final equilibrium temperature (in ∘ C ) when the same iron block initially at 110. 0 ∘ C is transferred into 400 mL of glycerol initially at 28. 0 ∘ C . For glycerol: Density = 1.25 g mL − 1 , Cp = 215 J mol − 1 K − 1 Assume specific heat capacity of water = 4.2 J g -1 K -1

TestHub · Accepted Answer

Heat gained by water: q = 1100 ( 4.2 ) ( 40 − 25 ) = 69300 J Heat lost by iron: 69300 = 1500 C F e ​ ( 140 − 40 ) = 1500 C F e ​ ( 100 ) C Fe ​ = 1500 × 100 69300 ​ = 0.462 Jg − 1 K − 1 Mass of glycerol: m = 400 × 1.25 = 500 g Moles ( M = 92 ): n = 92 500 ​ = 5.4348 mol Heat capacity of glycerol: C = n Cp ​ = 5.4348 × 215 = 1168.48 J / K Heat balance 1500 ( 0.462 ) ( 110 − T f ​ ) = 1168.48 ( T f ​ − 28 ) T f ​ = 58.5 2 ∘ C

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