# Determine the Specific Heat Capacity

• Psywing
In summary, the student attempted to find the Specific Heat Capacity of rubber by using a calorimeter, and found that the temp. change was 95-32. They then attempted to find the temp. change of the rubber, water, and calorimeter by using the mixing method, and found that the rubber had the highest S.H.C.
Psywing
Hello, everyone.

## Homework Statement

Now, I recently did an experiment in school to determine the Specific Heat Capacity of rubber using a calorimeter. We're using the mixing method to figure it out.
The mass of the calorimeter, rubber and water was measured. And the S.H.C is already known for the water and calorimeter. The temperatures are as follows:

Temp. of cold water = 26°C
Temp. of hot object = 95°C
Temp. of mixture = 32°C

## Homework Equations

Mo co θo = Mc cc θc + Mw cw θw

0 = object
c = calorimeter
w = water

## The Attempt at a Solution

The problem I'm having is finding the temp. change. This is what I did.
To find the temp. change of the rubber = 26 + 32 = 58. And from that 95 - 58 = 37.

To find the temp. change of the calorimeter = 95 - 32 = 63.

To find the temp. change of the water = 95 - 26.

I'd like to know if this is accurate because, when I do the calculations I get a really weird figure for the S.H.C of the rubber.

I'm confused--what was hot before mixing and cold before mixing. I'm assuming the rubber and calorimeter were both hot and the water cold.

My apologies, for not stating it properly.

The temp. of the mixture is the temp. of the heated rubber inside the calorimeter filled with water.

The temp. of the cold water is the temp. of the water inside the calorimeter alone.

The temp. of the hot object is the temp. of the heated rubber in a beaker of boiling water.

very well. So the temp difference between the rubber before and after is 95-32. The rise in temp of both the water and calorimiter by adding the rubber is 6 degrees.
Try those and see what you get. The signs of the temp changes should both be positive BTW.

Okay, I understand. Reading through your post made me realize that what you said makes more sense than what I did. There is no need to use the 26°C for the calculation of the temp. change of the rubber, right? The 32°C already has the 26°C included.

And, there's no need to use the 95°C for the calculation of the temp. change of the water and calorimeter because, the 32°C already has the 95°C included.

Is the way I stated it correct? I just want to know if I understand the logic behind it so I won't have any problems in the future.

You're interested in the heat gain on the one hand, so that's final temp minus initial temp for the water/calorimeter combo and heat loss on the other hand, which is again the initial temp minus the final. Heat is neither created or destroyed if we are careful so the heat loss (temp times mass times capacity) of the rubber is equal to the heat gain of the H20/calorimeter. That help. What value did you get by the way>

Oh, okay. I see. Many thanks, for that explanation.
I got 2162 J/kg K. The answer was rounded off. Was I supposed to convert the degrees Celsius to Kelvins before I made the calculations?
Because that's what I did.

Degrees K is customary but the caclulation should be the same. The result falls squarely in the range cited for India rubber.

Is that right? Well, I have no idea what type of rubber it is. So, I guess I'll stick with that answer. On Monday, I'll see if it's correct or at least close to being correct from my lecturer.

## 1. What is specific heat capacity and why is it important?

Specific heat capacity, also known as specific heat, is the amount of heat energy required to raise the temperature of 1 gram of a substance by 1 degree Celsius. It is an important physical property of a substance as it helps to determine how much heat energy is needed to raise or lower its temperature. This information is useful in various scientific fields, such as thermodynamics, chemistry, and engineering.

## 2. How is specific heat capacity measured?

Specific heat capacity is typically measured using a calorimeter, which is a device that can accurately measure the amount of heat absorbed or released by a substance. The substance is placed in the calorimeter and heated or cooled to a known temperature. The change in temperature, along with the mass of the substance, is then used to calculate the specific heat capacity using the formula q=mcΔT, where q is the amount of heat, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

## 3. What factors can affect the specific heat capacity of a substance?

The specific heat capacity of a substance can be affected by various factors, including its chemical composition, temperature, and phase (solid, liquid, or gas). Different substances have different specific heat capacities, and these values can also change with temperature. Additionally, the state of matter also plays a role, as the specific heat capacity of a substance can differ between its solid, liquid, and gas phases.

## 4. How does specific heat capacity relate to heat transfer?

Specific heat capacity is closely related to heat transfer, as it determines how much heat energy is needed to raise or lower the temperature of a substance. This information is crucial in understanding and predicting heat transfer processes, such as conduction, convection, and radiation. For example, substances with high specific heat capacities require more heat energy to raise their temperature, making them good insulators and less prone to temperature changes.

## 5. How is specific heat capacity used in real-world applications?

Specific heat capacity has many practical applications, such as in cooking, heating and cooling systems, and materials science. In cooking, specific heat capacity is used to determine the cooking time and temperature needed to cook different foods. In heating and cooling systems, specific heat capacity is used to calculate the amount of energy needed to heat or cool a building. In materials science, specific heat capacity is important in the design and development of new materials for various applications, such as thermal insulation and heat storage.

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