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.

Answers

root word, prefixes, and suffixes

v=ir = 3x15 =45 v

Explanation:

Given

Heat transfer Q=1.6\times 10^5\ J

initial Temperature T_i=17.5^{\circ}\approx 290.5\ K

Entropy change dS=485\ J/K

The expression for entropy is given by

dQ=TdS

T=\frac{dQ}{dS}

T=\frac{1.6\times 10^5}{485}

T=329.89\ K

Temperature can be written as average of initial and final temperature

T=\frac{T_i+T_f}{2}

329.89=\frac{T_f+290.5}{2}

T_f=659.78-290.5

T_f=369.28\ K



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