# Chemistry

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A 26.7 g sample of ethylene glycol, a car radiator coolant, loses 668 J of heat. What was the initial temperature of ethylene glycol if the final temperature is 32.5°C (c of ethylene glycol = 2.42 J/gK)?

• Chemistry -

q = mass e.g. x specific heat e.g. x (Tfinal-Tinitial)

• Chemistry -

668= 26.7(2.42)(X-32.5) ?

• Chemistry -

If Tfinal is 32.5 and Tinitial is X, it appears you have reversed the numbers. I have
-668 = 26.7(2.42)(32.5-X)
However, you reversed the sign on Tfinal-Tinitial AND you used a + instead of a - sign for heat lost of -668 so the sign turns out ok as well as the number.

• Chemistry -

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