How Does Heat Release Affect Water Temperature in Thermochemical Reactions?

[Signifier]

I know this is a physics forum, but this problem is mainly physics, and exceedingly simple:

Homework Statement

A certain reaction releases about 90 kJ of heat when stoichiometric amounts of reactants react. If the reaction goes to completion in 100 g of water whose initial temperature is 20 deg C, what is the final temperature of the water?

Homework Equations

q = mc(delta T), I believe, so delta T = q / (mc)

The Attempt at a Solution

I don't know what I'm doing wrong, but if I plug the 400 kJ = 400000 J into q and the 100 g mass into m and 4.184 J / (mol deg C) into c, I get an insane change in temperature around 215 degrees. I am obviously forgetting some step or misunderstanding something... I've always had trouble with thermodynamics stuff like this. Any help would be greatly appreciated.. thank you!

 

[Kurdt]

Firstly should q not be 90kJ. Secondly the units of c that you provide are J/(g K). Other than that your calculations seem fine.

 

[m2003]

I would first like to clarify that thermochemistry is a branch of physical chemistry, which combines principles from both physics and chemistry. Therefore, it is appropriate to ask thermochemistry questions in a physics forum.

Now, let's address the problem at hand. The equation you have used, q = mc(delta T), is correct. However, there are a few things to consider in order to arrive at the correct solution.

Firstly, it is important to note that the specific heat capacity of water is not a constant value. It varies with temperature. Therefore, the value of 4.184 J/(g deg C) that you have used is only valid for a specific temperature (usually 25 degrees Celsius). In order to get a more accurate answer, you can use the average specific heat capacity of water over the temperature range you are working with.

Secondly, the value of 90 kJ that is given in the problem is the total heat released by the reaction. This heat is not only used to raise the temperature of the water, but also to overcome the heat capacity of the container and any other surroundings. Therefore, the total heat used to raise the temperature of the water is less than 90 kJ.

Taking these factors into consideration, the correct approach would be to use the equation q = mc(delta T), where q is the total heat used to raise the temperature of the water, m is the mass of water, c is the average specific heat capacity of water, and delta T is the change in temperature.

I hope this helps you arrive at the correct solution. If you are still having trouble, I would recommend seeking assistance from a tutor or your professor. Thermochemistry can be a tricky subject, but with practice and understanding of the concepts, you will be able to solve problems like this easily.

 

[kyle8921]

I would first like to acknowledge the effort and attempt made by the individual to solve this problem. It is great to see someone actively seeking help and trying to understand a concept.

In this problem, we are dealing with thermochemistry, which is the study of the interconversion of heat and other forms of energy during chemical reactions. The equation used to calculate the change in temperature (delta T) is correct, but there are a few things that need to be considered.

Firstly, when calculating the heat released (q), we need to use the negative value of 90 kJ, as it is a exothermic reaction (heat is being released). So, q = -90 kJ.

Secondly, the specific heat capacity (c) of water is 4.184 J/(g °C), not 4.184 J/(mol °C). Therefore, we need to use 4.184 J/(g °C) as the value for c.

Finally, when we calculate the change in temperature (delta T), we need to use the mass of water in grams, not kilograms. So, the mass (m) should be 100 g, not 100 kg.

Using all these values, we can calculate the change in temperature as follows:

delta T = q / (mc)
= (-90 kJ) / (100 g x 4.184 J/(g °C))
= -215.5 °C

Since the change in temperature is negative, it means that the final temperature of the water will be lower than the initial temperature. Therefore, the final temperature of the water will be:

Tf = 20 °C - 215.5 °C
= -195.5 °C

This result may seem strange, but it is important to remember that we are dealing with a theoretical scenario and not real-life conditions. In reality, the water would not reach such a low temperature as it would start to freeze before that.

I hope this helps to clarify the concept and solve the problem. Keep up the good work!