Question

. 1.0kg of ice at 0oC is added to 10kg of water in a 2.0kg insulated aluminum container. The water and aluminum are both initially at 20oC. Calculate the final equilibrium temperature. (Lf ice = 3.3x105J/kg), (cAl = 900J/kgoC) In addition find the following 9. Calculate the heat needed to completely melt a 14kg bar of gold initially at 20oC. (Tmelt Au = 1064oC), ( cAu = 129J/kgoC), (Lf Au = 6.3x104J/kg)

Answer 1

heat required to melt the ice to water = Qreq = Mice*Lfice

Qreq = 1*3.3*10^5 = 3.3*10^5 J

heat lost by water + Al in melting ice Qlost = (Mw*Cw + MAl*CAl)*dT

Qlost = ((10*4186)+(2*900))*20 = 8.732*10^5 J

Qlost > Qreq

all the ice will melt and the temperatre rises to t

(Mw*Cw + MAl*CAl)*(20-t) = Mice*( Lfice + Cw*t)

((10*4186)+(2*900))*(20-t) = 1*( 3.3*10^5 + (4186*t))

equilibrium temperature t = 11.35 degrees celsius

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heat needed = heat req to raise the temp to 1064 + heat req to melt at 1064

Q = MAu*(CAu*dT + LfAu)

Q = 14*((129*(1064-20)) + (6.3*10^4))

Q = 2767464 J   <<<<+============ANSWER