Question

rn an experiment, 85.0 g of ali min um with a specic heat of gno 1 k K) at 97.n°C ls m oed with 34.0 9 nr wo er with a specific heat of 4185 נ/kg-K at 14.0 C wth the mixture thermally s lated a what is the equilibrium temperature? What are the entrap changes of b) the al minu (c) the water, and (d) the aluminum-water systeı? (a) Nunber (b) Number (c)Numter (d) Number Units Units

Answer 1

a) we know the formula

Q=mcdT

so here we have

Qw+Qa=0

Qw=-Qa

(mcdT)al=(mcdT)water

-900*0.085*(Tf-370.15)=4186*0.034*(Tf-287.15)

-76.5Tf+28316.475=142.324Tf-40868.3366

Tf=316K or 43⁰C

b)change in entropy=$\int_{i}^{f}$dQ/T

so dS=$\int_{i}^{f}$mcdT=mcln(Tf/Ti)=900*0.085*ln(316/370)=-12J/K

so the answer is -12 J/K or -12.1J/K

c) using the eqn from part b

ds=mcln(Tf/Ti)=0.034*4186*ln(316/287)=13.7 J/K

so the answer is 13.7 J/K or 14 J/K

d) entropy of aluminium water system=(-12+13.7)J/K=1.7J/K

so the answer is 1.7J/K or 2 J/K or 1.6 J/K or 1.64 J/K.