Ethylene glycol (antifreeze) has a specific heat of 2.42 J/(g*K). Calculate q when 3.65 kg of ethylene glycol cools from 240 degrees C to 78 C
q = mass e.g. x specific heat e.g. x (Tfinal-Tinitial) EXCEPT this is just part of it if the ethylene glycol goes through a phase change. Do you have a melting point and boiling point given?
no there is no melting or boliling point given.
I note here (link provided) that the boiling point of ethylene glycol is about 197 C which means that the molecule would be in the vapor state at 240 C (if it didn't decompose). So the person making up this problem failed to take that into account. There would be mass x specific heat vapor x (197-240) and there would be mass x deltaHvap when it condenses at 197. Both of these amounts would be in addition to what I wrote in the first response.
http://en.wikipedia.org/wiki/Ethylene_glycol
To calculate the amount of heat transferred, you can use the equation:
q = m * c * ΔT
where:
q is the amount of heat transferred (in Joules),
m is the mass of the substance (in grams),
c is the specific heat capacity of the substance (in J/(g*K)), and
ΔT is the change in temperature (in Kelvin).
First, convert the mass of ethylene glycol from kilograms to grams:
m = 3.65 kg * 1000 g/kg = 3650 g
Next, calculate the change in temperature:
ΔT = (240 °C - 78 °C) = 162 °C
Remember to convert this change in temperature to Kelvin by adding 273.15:
ΔT = 162 °C + 273.15 = 435.15 K
Finally, substitute the values into the equation:
q = 3650 g * 2.42 J/(g*K) * 435.15 K
Calculate the result using a calculator:
q ≈ 3597677.3 Joules
Therefore, the amount of heat transferred when 3.65 kg of ethylene glycol cools from 240 °C to 78 °C is approximately 3,597,677.3 Joules.