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When 1.60 × 10 5 J 1.60×105 J of heat transfer occurs into a meat pie initially at 17.5 °C , 17.5 °C, its entropy increases by 485 J / K . 485 J/K. Estimate the final temperature of the pie.

Respuesta :

Answer:

Explanation:

Given

Heat transfer [tex]Q=1.6\times 10^5\ J[/tex]

initial Temperature [tex]T_i=17.5^{\circ}\approx 290.5\ K[/tex]

Entropy change [tex]dS=485\ J/K[/tex]

The expression for entropy is given by

[tex]dQ=TdS[/tex]

[tex]T=\frac{dQ}{dS}[/tex]

[tex]T=\frac{1.6\times 10^5}{485}[/tex]

[tex]T=329.89\ K[/tex]

Temperature can be written as average of initial and final temperature

[tex]T=\frac{T_i+T_f}{2}[/tex]

[tex]329.89=\frac{T_f+290.5}{2}[/tex]

[tex]T_f=659.78-290.5[/tex]

[tex]T_f=369.28\ K[/tex]

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