A 50-kg copper block initially at 140°c is dropped into an insulated tank that contains 90 l of water at 10°c. Determine the final equilibrium temperature and the total entropy change for this process.

Answers

Answer 1

Answer:

[tex]T_f=24.71[/tex]

Explanation:

From the question we are told that:

Mass of block [tex]m=50[/tex]

Temperature of block [tex]T_b =140 \textdegree C[/tex]

Volume of water [tex]V= 90L[/tex]

Temperature of water [tex]T_w=10 \textdegree C[/tex]

Density of water [tex]\rho=1000kg/m^3[/tex]

Specific heat of water [tex]C_w=4.18KJ/kg-k[/tex]

Specific heat of copper [tex]C_p=0.96KJ/kg-k[/tex]

 

Generally the equation for equilibrium stage is mathematically given by

[tex]mC_p(T_b-T_f)=\rho*VV*c(T_f-T_w)[/tex]

[tex]50*0.96(140-T_f)=1000*90*10^-3*c_w(T_f-10)[/tex]

[tex]48(140-T_f)=376.2(T_f-10)[/tex]

[tex]140-T_f=7.8375(T_f-10)[/tex]

[tex]140-T_f=7.8375T_f-78.375[/tex]

[tex]-8.8375T_f=-218.375[/tex]

[tex]T_f=\frac{-218.375}{-8.8375}[/tex]

[tex]T_f=\frac{-218.375}{-8.8375}[/tex]

[tex]T_f=24.71[/tex]


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