A 80-lb aluminum bar, initially at Ta = 150°F, is placed in a tank together with

Question

A 80-lb aluminum bar, initially at Ta = 150°F, is placed in a tank together with 190 lb of liquid water, initially at Tv = 70°F, and allowed to achieve thermal equilibrium. The aluminum bar and water can be modeled as incompressible with specific heats ca0.216 Btu/lb.°R and cw 0.998 Btu/lb.°R, respectively. Consider the aluminum bar and water as the system and ignore heat transfer between the system and its surroundings Determine the final temperature Tf, in OF and the amount of entropy produced within the tank, in Btu/°R.

Explanation / Answer

Solution:

Heat Loss = Heat Gain

mw . cw . (Tf - T1w) = mb . cb . (T1b - Tf ?)

190 * 0.998 * (Tf - 70) = 80 * 0.216 * (150 - Tf )

So,   Tf = 76.68o F

Amount of Entropy produced:

dS = dSw + dSb

dS = mw . cw . ln(Tf /T1w) + mb . cb . ln(Tf /T1b)

dS = 17.28 - 11.59 = 5.69 Btu/oR

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