A 2.5 kg aluminum container is heated to 80degreeC. At this temperature it can h

Question

A 2.5 kg aluminum container is heated to 80degreeC. At this temperature it can hold 3.000 Times 10^-3 m^3 of fluid. A student pours 3.000 Times 10^-3 m^3 of water at 5degreeC into the can The heat capacity of aluminum is 900 J/kg degree C and tlie lieat capacity of water is 4190 J/kg degree C. (a) What is its final temperature assuming that no heat escapes the system? (b) The volume expansion coefficient for aluminum is 7.2 x 10^-5degreeC^-1, and the volume expansion coefficient for water is 1.5 x 10^-4degreeC^-1. As a result, as the canister cools it shrinks, while the warming water expands Estimate the mass of water which spills out of the vessel when the system reaches equilibrium. Assume that the density of water is exactly 1 g/cm^3 for the entire expermient.

Explanation / Answer

Mal = 2.5 Kg
Tal = 80
Val = 3*10^-3m^3
Cal = 900
Vw = 3*10^-3
Tw = 5
Cw = 4190
let the final temperature be T

Mw*Cw*(T -5) = Mal*Cal*(80-T)
Mw = (rho)w * Vw = 3 kg
3*4190(T-5) = 2.5*900*(80-T)
12570T - 62850 = -2250T + 180000
T = 16.38 deg

Kal = 7.2*10^-5
Kw = 1.5*10^-4

New Volume of aluminum after contracting = Val - Val*Kal*(80-16.38) = 3*10^-3[1-7.2*10^-5*63.62] = 2.986*10^-3
New Volume of water after expanding = Vw + Vw*Kw*(16.38 - 5) = 3*10^-3[1+1.5*10^-4*11.38] = 3.005*10^-3
Net water spilled = Vw - Val = 0.0191*10^-3 m^3

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