Please help me with my THERMAL PROPERTIES OF MATTER

In summary, the question is asking for the specific heat capacity of the object when it is placed in water. Using the formula for thermal energy, and considering that no heat is lost to the surroundings and the heat absorbed by the beaker is negligible, we can equate the energy lost by the water to the energy gained by the object to calculate the specific heat capacity of the object.
  • #1
dansics
5
0


The question I'm currently working on:
When 0.6 kg of an object at 6 degree Celsius is put inside a beaker of water of mass 1.2Kg at 28 degree Celsius, the final temperature of water is 26 degree Celsius. Assuming no heat lost to the surroundings and the heat absorbed by the beaker is negligible, calculate the specific heat capacity of the object. (Specific heat capacity of water is 4200 J KG-1 K-1)

So far.. MY solution is:

Thermal energy, Q= specific heat capacity of water x mass of object and water x the change in temperature
which is: 4200 x 1.8 x (-8) = -60480

I am really clueless right now.. Hope you guys are able to help and guide me with this one..
 
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  • #2
Nearly correct,
The change in energy of the water is:
Q1 = specific heat of water * mass of WATER * change in temperature of water

Then you can use this with:
Q2 = specific heat of object * mass of OBJECT * change in temperature of object

Since no heat is lost, Q1 lost by water equals the Q2 gained by object.
 
  • #3


Hello! It seems like you are on the right track with your solution. To find the specific heat capacity of the object, we can use the formula Q = mcΔT, where Q is the thermal energy, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

In this case, we know the mass of the water (1.2 kg), the specific heat capacity of water (4200 J kg-1 K-1), and the change in temperature (8 degrees Celsius). We can plug these values into the formula and solve for the specific heat capacity of the object.

Q = mcΔT
-60480 = (0.6 kg + 1.2 kg)(c)(-8)
-60480 = 1.8 c (-8)
c = -60480 / (-14.4)
c = 4200 J kg-1 K-1

So, the specific heat capacity of the object is also 4200 J kg-1 K-1. This means that the object has the same ability to absorb and release thermal energy as water does. I hope this helps guide you in your calculations. Keep up the good work!
 

1. What are the basic concepts of thermal properties of matter?

Thermal properties of matter refer to the physical characteristics and behaviors of matter when it is exposed to heat or changes in temperature. The three main concepts are thermal expansion, specific heat, and thermal conductivity.

2. How does thermal expansion affect matter?

Thermal expansion is the tendency of matter to increase in size when heated and decrease in size when cooled. This is due to the particles in matter gaining or losing kinetic energy, causing them to move apart or closer together. This can be seen in everyday objects such as bridges expanding on hot days and contracting on cold days.

3. What is specific heat and why is it important?

Specific heat is the amount of heat required to raise the temperature of a substance by 1 degree Celsius. It is an important property because it determines how much heat is needed to change the temperature of a substance. Different substances have different specific heats, which can affect their ability to store or release heat.

4. How does thermal conductivity affect the transfer of heat?

Thermal conductivity is the measure of a material's ability to conduct heat. Materials with high thermal conductivity, such as metals, can transfer heat quickly, while materials with low thermal conductivity, such as insulators, are better at retaining heat. This property is important in understanding how heat is transferred through different materials and can impact the efficiency of thermal insulation and heat exchange systems.

5. What are some real-world applications of thermal properties of matter?

Understanding thermal properties of matter has many practical applications, such as in building and construction materials, cooking and food preservation, and energy production. It also plays a crucial role in fields such as materials science, engineering, and environmental studies. Overall, knowledge of thermal properties of matter is essential in designing and improving various technologies and processes in our daily lives.

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