1) 3.0kg of water at 34?C is mixed with 3.0kg of water at 70?C in a well-insulat

Question

1) 3.0kg of water at 34?C is mixed with 3.0kg of water at 70?C in a well-insulated container.Estimate the net change in entropy of the system.

Express your answer to two significant figures and include the appropriate units.

2) Which of the following possibilities could increase the efficiency of a heat engine or an internal combustion engine?

Select 2 or more that apply.

Increase the temperature of the hot part of the system and reduce the temperature of the exhaust. Increase the temperatures of both the hot part and the exhaust part of the system by the same amount. Decrease the temperatures of both the hot part and the exhaust part of the system by the same amount. Decrease the temperature of the hot part and increase the temperature of the exhaust part by the same amount. None of the above; only redesigning the engine or using better gas could improve the engines efficiency.

Explanation / Answer

1)

here both liquid has the equal masses of at T1=34+273=307 K and T2=70+273=343 K are mixed,
let temperature of mixure be Tm,

then
heat lost =heat gain

m*S*(T1-Tm)=m*s*(Tm-T2)

Tm=(T1+T2)/2

=307+343/2

Tm=325 K

now
a)
chenge of entropy of 3kg of water at T1 k is converting into water at Tm C is,
dS1=m*S_water*ln(Tm/T1)

=3*4186*2.303*log(325/307)

=715.651 J/K

b)

chenge of entropy of 3kg of water at T2 k is converting into water at Tm C is,
dS2=m*S_water*ln(Tm/T2)

=3*4186*2.303*log(325/343)

=-677.064 J/K

the net change in entropy of the system is,
dS=dS1+dS2

=715.651-677.064

=38.587 J/K .......is answer

2)

efficiency =1-T2/T1

from the above equation we could increase the efficiency of a heat engine by

Increase the temperature of the hot part of the system and reduce the temperature of the exhaust.

but practicaly above condition is not possible easily so, the better option is,

None of the above; only redesigning the engine or using better gas could improve the engines efficiency.

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