Several ice cubes (rho_i = 0.9167 g/cm^3) of total volume V_i = 265 cm^3 and tem

Question

Several ice cubes (rho_i = 0.9167 g/cm^3) of total volume V_i = 265 cm^3 and temperature 273.15 K (0.000 degree C) are put into a thermos containing V_t = 560 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 the magnitude of the total heat transferred Q_T in J from the tea to the ice cubes.

Explanation / Answer

(A) Q = m Lf

m = volume x density = 265 x 0.9167

m = 242.93 g

Q = 242.93 x 334 = 81137.12 J

(B) energy absorbed by water + 81137.12 = energy released by tea

(242.93 x 4.186 x T ) + 81137.12 = 560 x 4.186 x (40.15 - T)

1014.27 T + 81137.12 = 94118.02 - 2344.16T

T = 3.87 deg C

(C) Q = 560 x 4.186 x (40.15 - 3.87)

= 85057.5 J

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