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=mc∆t
Q = heat energy
m = mass
c = specific heat (of ethylene glycol)
∆t= change in temperature (= final - initial temp)
To solve this problem, we can use the formula for heat transfer:
Q = m * c * ΔT
where:
Q is the amount of heat transferred
m is the mass of the substance
c is the specific heat capacity of the substance
ΔT is the change in temperature
In this case, we are given:
m = 26.7 g (mass of ethylene glycol)
Q = -668 J (negative because heat is lost)
c = 2.42 J/gK (specific heat capacity of ethylene glycol)
ΔT = final temperature - initial temperature = 32.5°C - initial temperature
Now, let's substitute these values into the formula and solve for the initial temperature:
-668 J = 26.7 g * 2.42 J/gK * (32.5°C - initial temperature)
To simplify the equation, convert the temperature from Celsius to Kelvin:
32.5°C = 32.5 K + 273.15 K = 305.65 K
-668 J = 26.7 g * 2.42 J/gK * (305.65 K - initial temperature)
Simplify further:
-668 J = 64.554 J * (305.65 K - initial temperature)
Divide both sides by 64.554 J:
-668 J / 64.554 J = 305.65 K - initial temperature
Simplify:
-10.35 = 305.65 K - initial temperature
Rearrange to solve for the initial temperature:
initial temperature = 305.65 K - (-10.35)
initial temperature = 305.65 K + 10.35
initial temperature ≈ 316 K
Therefore, the initial temperature of ethylene glycol was approximately 316 K.