Calculate the amount of heat needed to boil 47.3 g of octane, beginning from a temperature of 1.2 C . Be sure your answer has a unit symbol and the correct number of significant digits.

Answer 1

For this problem, we will be using the q = m*c*(delta T) equation, which is also known as the Specific Heat formula.

Q is the amount of heat added to a substance (in Joules), m is the mass of the substance (in kg), c is the specific heat of that particular substance (which varies for each different substance we work with) (in units of J/kg * K) and delta T is the change in temperature (in the units of Kelvin).

We need to know the specific heat of octane for solving this problem, and from a Google search, I find that the specific heat of octane is 2400 J/kg * K

Next, I convert the 47.3g of octane we have into kg of octane. This is done by dividing by a thousand.

47.3g * (1 kg/ 1000g) = 0.0473 kg of octane

For this problem, I also need the boiling point of octane. From a Google search I find that the boiling point of octane is 398.8 K. The initial temperature of octane in our problem is 1.2 C. I will have to convert this temperature, which is given to us in Celsius, into Kelvins by adding 273.15.

Kelvin = Celsius + 273.15

Kelvin = 1.2 C + 273.15

K = 274.35

Delta T is the change in temperature (final temperature minus the initial temperature). My delta T will be 398.8 K - 274.35 K.

So now, I have all of the information that I need to plug into my specific heat equation and to finish my problem.

q = m * c * (delta T)

q = (0.0473 kg) * (2400 J/kg * K) * (398.8 K - 274.35 K)

q = 14,127.564 J

We have to round this answer with the correct number of significant digits. Since our specific heat (c) has only 2 significant digits, our answer should only have two as well. So our answer will be 14,000 Joules or you can say 14 kJ (kilojoules).

Answer 2

We need to know several material facts before answering this. What are the freezing and boiling points of octane. What are the thermochemical constants for octane, i.e. specific heat of solid, liquid and gaseous octane? What is the ∆Hf and ∆Hvap of octane.

So, looking up some of these values, I find

melting point = ~ -55ºC

boiling point = 125ºC

So, at 1.2ºC, the octane will be a liquid. Therefore the only other piece of info we need is the specific heat for liquid octane. I found that to be ~2.25 J/gº

Now, to answer the question:

use q = mC∆T

q = heat = ?

m = mass = 47.3 g

C = specific heat = 2.25 J/gº

∆T= change in temperature =125º - 1.2º = 123.8º

q = (47.3 g)(2.25 J/gº)(123.8º)

q = 13,175 J = 13.2 kJ = 13 kJ (2 sig. figs. based on 1.2ºC having 2 s.f.)