A student mixes .075 kg of an unknown substance at 96.5°C with .075 kg of water at 25.0°C. If the final temperature of the system is 31.15°C, what is the specific heat capacity of the substance?

The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.

Explanation:

We notice that the student is mixing a substance with a high temperature and another substance with a low temperature, where the first release heat to the latter one until thermal equilibrium is reached. By the First Law of Thermodynamics and assuming that the entire system has no energy interactions with the surroundings, we get the following model:

(1)

Where and are the changes in internal energy for the unknown substance and water, measured in joules.

By definition of internal energy, we expand the equation above now:

(2)

Where:

, - Masses of the unknown substance and water, measured in kilograms.

, - Specific heats of the unknown substance and water, measured in joules per kilogram-degree Celsius.

, - Initial and final temperatures of the unknown substance, measured in degrees Celsius.

, - Initial and final temperatures of water, measured in degrees Celsius.

Then, we clear the specific heat of the unknown substance:

If we know that , , , and , then the heat capacity of the unknown substance is:

The specific heat of the substance is 393.939 joules per kilogram-degree Celsius.

dang that balls went far

1 is a answer of a question