A 27.6 g sample of ethylene glycol, a car radiator coolant, loses 680. J of heat.?

Question

What was the initial temperature of ethylene glycol if the final temperature is 32.5°C (c of ethylene glycol = 2.42 J/g·K)?

Answer 1

680 J = (27.6 g) x (2.42 J/g·K) x (32.5- T)
32.5 - T = 10.18
T= 32.5 - 10.18
T= 22.32 °C

Answer 2

you have the formula for the lossed heat

Q = - mcdt here m=27.6 c=2.42*J/g-K Q=680

so dt = -680/(2.42*27.6) = -10.18°C

So the coolant lost 10.18c and initial temperature was

32.5+10.18= 42.68°C