A 49.3 g sample of ethylene glycol, a car radiator coolant, loses 624. 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)?
624= 49.3 x 2.42(T-32.50)
T = 37.73 = 38degrees
is this correct?
thank you.
1 answer
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1 answer