An aluminum MAl (g)cup contains Mw (g) of water and a MCu (g) copper stirrer, al

Question

An aluminum MAl (g)cup contains Mw (g) of water and a MCu (g) copper stirrer, all at T1 (°C). A MAg(g) sample of silver at an initial temperature of T2°C is placed in the water. The stirrer is used to stir the mixture until it reaches its final equilibrium temperature of Teq°C. a) Find an expression for the equilibrium temperatue Teq°C and b) the mass of the aluminum cup when Mw (g)= 220g, MCu (g)= 38g, T1 (°C)=85°C , MAg(g)= 390g, T2(°C)=87°C , Teq(°C)=32°C , MAl (g)=?, c) Include a detailed sketch

Explanation / Answer

let,

T1=85 oC

T2=87 oC

Teq=32 oC

mass of water mW=220g

mass of copper stirrer m_cu=38 g

mass of silver sample m_Ag=390g

mass of the aluminum cup is m_Al

a)

by using law of conservation of energy,

m_Al*C_Al*dT+m_w*C_w*dT+m_cu*C_cu*dT+m_Ag*C_Ag*dT=0

m_Al*C_Al*(T2-T)+m_w*C_w*(T2-T)+m_cu*C_cu*(T2-T)+m_Ag*C_Ag*(T1-T)=0

====>

equilibrium temperature T=((m_Al*C_Al+m_w*C_w+m_cu*C_cu)*T2+(m_Ag*C_Ag)*T1)/((m_Al*C_Al)+(m_w*C_w2)+(m_cu*C_cu)+(m_Ag*C_Ag))

b)

m_Al*C_Al*(T2-T)+m_w*C_w*(T2-T)+m_cu*C_cu*(T2-T)+m_Ag*C_Ag*(T1-T)=0

m_Al*0.9*(85-32)+220*4.186*(85-32)+38*0.387*(85-32)+390*0.235*(87-32)=0

===> m_Al=1145 g

mass of the Aluminum cup m_Al=1145.26 kg

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