I keep getting 292.18 K, but that is not right? Any help would be appreciated! S

Question

I keep getting 292.18 K, but that is not right? Any help would be appreciated!

Service ice cubes (rho_t = 0.9167 g/cm^3) of total volume V_t = 285 cm^3 and temperature 273.15 K (0.000 degree C are put into a thermos containing V_t = 580 cm^3 of tea at a temperature of 313.15 K, completely filling the thermos. The lid is then put on the thermos to close it. Assume that the density and the specific heat of the tea is the same as it is for fresh water (rho_w = 1.00 g/cm^3, c = 4186 J/kgK). Calculate the amount of heat energy Q_m in J needed to melt the ice cubes (L_f = 334 kJ/kg) Calculate the equilibrium temperature T_E in k of the final mixture of tea and water. calculate magnitude of the total heat transferred in Q_T in J from the tea to the ice cubes.

Explanation / Answer

mass of ice = density*volume = 0.9167*285 = 261.2595 g

heat energy Qm = mice*Lf = 261.2595*10^-3*334*10^3 = 87260.673 J

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part (b)

mass of water = 580*1 = 580 g = 0.58 kg

if all the ice melts to water at 0oc

heat trasfered from water = heat absorbed by ice

Mwater*swater*(TW-TE) = Mice*Lf + Mice*Swater(TE-Tice)

0.58*4186*(313.15-TE) = (261.2595*10^-3*334*10^3) + (261.2595*10^-3*4186*(TE-273.15))

TE = 275.95 K <<<------answer

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heat transferred = QT = Mwater*swater*(TW-TE) = 0.58*4186*(313.15-275.95) = 90317.14 J

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