How would you calculate the answers to B and C? A= 90320 Several ice cubes (Q_i-

Question

How would you calculate the answers to B and C?

A= 90320

Several ice cubes (Q_i- = 0.9167 g/cm^3) of total volume V_i = 295 cm^3 and temperature 273.15 K (0.000 degree C) are put into a thermos containing V_t = 590 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 (Q_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

mass of ice m = 0.9167 g/cm 3 x 295 cm 3

                     = 270.4265 g = 0.2704265 kg

Mass of tea M = 1 g/cm 3 x 590cm 3

                         = 590 g = 0.59 kg

Initila tempearture of ice t = 0 o C

Initial temperature of tea t ' = 313.15 K -273.15 K = 40 o C

heat lost by tea = heat gain by water

MC(t ' - T ) = mL +mCT

0.59x4186(40 -T ) = (0.2704265x334000)+(0.2704265x4186T)

98789.6 -2469.74 T = 90322.451 + 1132 T

3601.74 T = 8467.149

            T = 2.35 o C

               =2.35 + 273.15

               = 275.5 K

(c). Total transferred Q T = 0.59x4186(40 -2.35)

                                    = 92985.7 J

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