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.
w
Well, you're really heating things up with this question! But let me help you out and check your calculations.
You're correct that the heat lost (624 J) is equal to the mass (49.3 g) multiplied by the specific heat capacity (2.42 J/gK) multiplied by the change in temperature (T - 32.5). So, we have:
624 = 49.3 x 2.42 x (T - 32.5)
Let's do some math to solve for T:
624 = 119.306 x (T - 32.5)
624 = 119.306T - 3878.185
119.306T = 4502.185
T ≈ 37.72°C
So, it seems like you made a slight rounding error in your final answer. The initial temperature is approximately 37.72°C or 38°C if we round it. Great job overall, and always remember to keep your cool when dealing with thermodynamics!
To determine the initial temperature of ethylene glycol, we can use the equation:
Q = m * c * ΔT
where Q is the heat loss (624 J), m is the mass of ethylene glycol (49.3 g), c is the specific heat capacity of ethylene glycol (2.42 J/gK), and ΔT is the change in temperature (final temperature - initial temperature).
Rewriting the equation, we have:
624 J = 49.3 g * 2.42 J/gK * (T - 32.5°C)
Now we can solve for the initial temperature T:
624 J = 119.246 gJ/K * (T - 32.5°C)
624 J = 119.246 gJ/K * T - 3876.685 gJ
624 J + 3876.685 gJ = 119.246 gJ/K * T
4500.685 gJ = 119.246 gJ/K * T
T = 4500.685 gJ / 119.246 gJ/K
T ≈ 37.73987 ≈ 38°C
So the initial temperature of ethylene glycol was approximately 38°C. Your calculation is correct.
To determine if your answer is correct, let's go through the calculation step-by-step:
The formula for calculating the heat transferred is given by:
q = mcΔT
Where:
q is the heat transferred
m is the mass of the substance in grams (49.3 g in this case)
c is the specific heat capacity of the substance (2.42 J/gK in this case)
ΔT is the change in temperature, which is the final temperature minus the initial temperature (32.5°C - T)
So, the equation becomes:
624 J = (49.3 g) x (2.42 J/gK) x (32.5°C - T)
Now let's solve for T:
624 J = 119.306 J/K x (32.5°C - T)
624 J = 3877.96 - 119.306 T
Rearranging the equation to solve for T:
119.306 T = 3877.96 - 624
119.306 T = 3253.96
T = 3253.96 / 119.306
T ≈ 27.27°C
Therefore, based on the calculation, the initial temperature of ethylene glycol is approximately 27.27°C, not 38 degrees as you initially suggested.