Question #142262

A copper calorimeter that weighs 300g contains 200g Oil initially at 293.15K. 180g of Aluminium at 300°C is added to the oil. What will be the final temperature of the system after equilibrium is established?

Specific heat capacity for Copper = 0.379J/g°C

Specific heat capacity for Aluminium= 0.921J/g°C

Specific heat capacity for Oil= 1.54 J/g°C


1
Expert's answer
2020-11-05T10:10:49-0500

mc=300gmo=200gTc=293.15K=20°CTo=20°Cma=180gTa=300°CTfinal=?m_c = 300g\\ m_o = 200g\\ T_c = 293.15K = 20°C\\ T_o = 20°C\\ m_a = 180g\\ T_a = 300°C\\ T_{final} = ?


cc=0.379 J/g°Cca=0.921 J/g°Cco=1.54 J/g°Cc_c = 0.379\ J/g°C\\ c_a = 0.921\ J/g°C\\ c_o = 1.54\ J/g°C



According to the Law of conservation of energy,

Heat gained by substance A = Heat lost by substance B


mccc(TfTc)+moco(TfTo)=maca(TaTf)300×0.379(Tf20)+200×1.54(Tf20)=180×0.921(300Tf)113.7(Tf20)+308(Tf20)=165.78(300Tf)(113.7+308)(Tf20)=49734165.78Tf421.7(Tf20)=49734165.78Tf421.7Tf8434=49734165.78Tf587.48Tf=58168Tf=99.01°C\rightarrow m_c c_c (T_f - T_c) + m_o c_o (T_f - T_o) = m_a c_a (T_a T_f)\\ \rightarrow 300×0.379(T_f -20) + 200×1.54(T_f-20) = 180×0.921(300-T_f)\\ \rightarrow 113.7(T_f-20) + 308(T_f-20) = 165.78(300-T_f)\\ \rightarrow (113.7+308)(T_f-20)= 49734-165.78T_f\\ \rightarrow 421.7(T_f-20)= 49734-165.78T_f\\ \rightarrow 421.7T_f - 8434=49734-165.78T_f\\ \rightarrow 587.48T_f = 58168\\ \rightarrow T_f = 99.01°C


\therefore The final temperature of the system after equilibrium is established will be 99.01°C99.01°C

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