(10%) Problem 9: m! = 1.3 kg of ice at 0°C and 7nw= 3.6 kg of water at Tw = 97 C

Question

(10%) Problem 9: m! = 1.3 kg of ice at 0°C and 7nw= 3.6 kg of water at Tw = 97 C are mixed together in an insulated container Water's specific heat and latent heat offusion are cw = 4.19x103 J/(kg·°C) and Lf= 3.34x 105 J kg, respectively > 50% Part (a) Enter an expression for the final, equilibrium temperature of the mixture, in terms of the defined quantities. Grade Summa Potential Submissions 0% 100% Attempts remaining: 10 (5% per attempt) detailed view Lf mw Tw PACE Submit Hint I give up! Hints: 2 % deduction per hint. Hints remaining: 4 Feedback: 0% deduction per feedback. 50% Part (b) Calculate the final equilibrium temperature of the mixture, in degrees Celsius

Explanation / Answer

From the principle of calorimetry, use Heat lost by water = Het gained by ice

Mass of water mw =3.6 kg

Specific heat of water = Sw = 4190 J/kg C

Tw be the initial temperature of water = 97 deg C

Let final resultant temperature be Tf = ?

Mass of ice = mi = 1.3 kg

Latent heat of ice=Li=3.34 e5 J/kg

So mw cw dT = miLi + mi Cw *dT1

mw*cw *(Tw-Tf) = mi Li + mi Cw *(Tf-Ti)

expressiin for Tf = (mw cw Tw) –(miLi) + (mi cw Ti)/(mw cw + mi Cw)

part B:

Tf = (3.6 * 4190* 97) –(1.3* 3.34*10^5) +(1.3 *4190*0)/((3.6+1.3)*4190)

Tf = 50.11 deg C

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