Asked by Brittany

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)?
Answered by DrBob222
q = mass e.g. x specific heat e.g. x (Tfinal-Tinitial)
Answered by Britt
668= 26.7(2.42)(X-32.5) ?
Answered by DrBob222
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.
Answered by Anonymous
what is the answer????????
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