Question

5.20L of an ethylene glycol/water mixture in a car radiator releases 3520 kJ of heat. If the final temperature of the ethylene glycol/water mixture is 25.0°C, what was the initial temperature of ethylene/glycol water mixture?

For the ethylene glycol/water mixture, c=3.140 J/g°C,

d=1.11g/mL

Answer 1

Applying Q = m x c x (T)   for the ethylene glycol/water mixture

Given : Q = - 3520 kJ ,    Note that the negative sign indicates that heat is released.

             m = volume x denisty (d)   = 5.20 L x 1.11g/mL = (5.20 x1000 ml) x 1.11 g/ml = 5772 g

              c = 3.140 J/g°C

Putting the values:

- 3520 kJ = (5772 g) x (3.140 J/g°C ) x T

T        = - (3520 x 10³)J /{(5772 g) x (3.140 J/g°C )}

             = - 194.22 °C

Note that final temperature (T_f) of the ethylene glycol/water mixture is 25.0°C

And T = T_f - T_i = 25.0°C - T_i = - 194.22 °C

hence, T_i = 25.0°C + 194.22 °C = 219.22 °C

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