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)?
q = mass e.g. x specific heat e.g. x (Tfinal-Tinitial)
668= 26.7(2.42)(X-32.5) ?
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.
what is the answer????????
To solve this problem, we can use the equation:
q = mcΔT
Where:
q is the heat absorbed or released
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature
We are given:
q = -668 J (negative because heat is lost)
m = 26.7 g
c = 2.42 J/gK
ΔT = Tf - Ti
We want to find the initial temperature (Ti).
Rearranging the equation, we have:
q = mcΔT
q = mc(Tf - Ti)
Substituting the known values, we have:
-668 J = (26.7 g)(2.42 J/gK)(32.5°C - Ti)
Now, let's solve for Ti:
-668 J = (26.7 g)(2.42 J/gK)(32.5°C - Ti)
-668 J = 64.914 J/°C(g)(32.5°C - Ti)
Dividing both sides by 64.914 J/°C(g):
-668 J / 64.914 J/°C(g) = 32.5°C - Ti
Simplifying:
-10.28 °C = 32.5°C - Ti
Rearranging the equation to solve for Ti:
Ti = 32.5°C + 10.28 °C
Ti = 42.78°C
Therefore, the initial temperature of ethylene glycol was approximately 42.78°C.