The highest theoretical possible temperature of water

Question

The molar heat of combustion of pentane (C₅H₁₂) to form carbon dioxide and water liquid is - 3509 kJ/mol. If 4.65 g of pentane is burned to heat the 2.00 L of water at 20.0°C, the highest theoretical possible temperature of the water is ab.c °C. 

Answer 1

First, we will determine the amount of heat that is released from the combustion of 4.65 g of pentane.

convert 4.65 g pentane to moles of pentane: 4.65 g x 1 mol / 74.1g = 0.06275 mols

heat released = 0.06275 mols x -3509 kJ/mol = -220.2 kJ (the - sign tells us heat is being released)

Now that we know how much heat was released from the combustion reaction, we can find the temperature of the water that absorbed that amount of heat.

The equation we use is q = mC∆T

q = heat = 220.2 kJ = 220200 J

m = mass of water = 2.00 L = 2000 g (assuming a density of water of 1 g / ml)

C = specific heat for water = 4.184 J/gº

∆T = change in temperature = ?

Solving for ∆T we have...

q = mC∆T

∆T = q / (m)(C) = 220200 J / (2000 g)(4.184 J/gº)

∆T = 26.3º

This is the CHANGE in temperature, and since the water is absorbing heat, the temperature of the water will rise. The final temperature is thus...

20.0º + 26.3º = 46.3ºC

Answer 2

Hello Judy! Since this is a thermochemistry problem, let’s use the thermodynamic equation for heat:

q = mc(Tf-Ti), where q is the heat absorbed or released in J, m is the mass of the substance in grams, c is the specific heat capacity of the substance in J/g-°C, and Tf-Ti is the change in temperature in either °C or K. Remember, the difference in temperature in K is the same as the difference in temperature in °C, so use what is more convenient. We will use °C since this is what the question calls for.

We are asked to find the final temperature of the WATER, therefore all of our variables must pertain to the WATER. We know the amount of water (2.00 L, since the density of water is 1000 g/L, then how much water do we have..?) and the initial temp (20.0°C). We also know the specific heat capacity of water, it is always 4.18 J/g-°C. We do not know the final temperature or the heat that is absorbed by the water. What can we do?

Well, since pentane and water are the only two species in the system, we can assume that all of the heat that is released from the combustion of pentane is absorbed by the water. Since we know the molar heat capacity of pentane (-3509 kJ/mol) and we know we have 4.65 g of pentane, then we can finding the heat released from the combustion of pentane once we convert the grams to moles (molar mass of pentane: 72.15 g/mol).

Two things to keep in mind for this step: 1) remember we need q to be in J in order to plug it into our first equation for the heat absorbed by the water. 2) The water is ABSORBING heat, while the combustion reaction is RELEASING heat. According to the first law of thermodynamics, the energy of this system must be conserved. Therefore:

q(water)= -q(pentane).

Once you find q for the water, plug it back into the first equation and you should be able to solve for Tf! I hope this helps!