Several ice cubes (Q_i = 0.9167 g/cm^3) of total volume V_i = 220 cm^3 and tempe

Question

Several ice cubes (Q_i = 0.9167 g/cm^3) of total volume V_i = 220 cm^3 and temperature 273.15 K (0.000degreeC) are put into a thermos containing V_i = 620 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/kg). Calculate the amount of heat energy A_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

part a:

mass of ice cube=volume*density=220*0.9167=201.67 grams=0.20167 kg

then energy required to melt the ice completely=mass of ice*latent heat of fusion=

0.20167 kg*334 kJ/kg=67.36 kJ

part b:

mass of tea=620*1=620 grams=0.62 kg

let final temperature be T K.

then heat lost by tea=heat required to melt the ice cubes+heat required to raise the temperature of
liquid water from 273.15 K to T K

==>0.62*4186*(313.15-T)=67.36*1000+0.20167*4186*(T-273.15)

==>0.62*4186*313.15-0.62*4186*T=67.36*1000+0.20167*4186*T-0.20167*4186*273.15

==>T=283.748 K

part c:

total heat transferred from tea to ice cube=0.62*4186*(313.15-T)=7.6308*10^4 J

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