a fish at a pressure of 1.10 atm has its bladder inflated to an initial volume o

Question

a fish at a pressure of 1.10 atm has its bladder inflated to an initial volume of 8.2mL. if the fish starts swimming horizontally, its temperature increases from 20.1C to 22.9C as a result of exertion.
A) since fish is at same pressure, how much work is done to by air in the swim bladder(mJ)?
(B) how much heat is gained by the air in the swim bladder(mJ)?
(C)if this amount of heat is lost by the fish , by how much will its temperature decrease(mK)?
mass of fish=5g & specific heat =3.5J/(g*K).

Explanation / Answer

 A) V1/T1=V2/T2 ,V2=8.28 mL , W=p x delta V=1.10 x 0.08 x 10^-3 = 0.088 mJ

B) Q= C x deltaT=3.53 x 2.8 = 9.8 J = 9.8 x 10^3 mJ

C) If this amnt of heat is lost,temperatur decrease will be 2.8 K .

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