Calculating Specific Heat Capacity

In summary, the specific heat capacity of the solid material can be calculated using the equation c = Q/(m*change in T), where Q is the energy provided, m is the mass of the material, and change in T is the change in temperature. Plugging in the values of 725 kJ of energy, 4.96 kg mass, and a change in temperature of 217.3 degrees Celsius, we get a specific heat capacity of 230 J/kg/K. However, the answer may need to be converted to kJ/kg/K and have less significant figures for it to be accepted as correct.
  • #1
struggtofunc
15
1

Homework Statement


A 4.96 kg piece of solid material is heated from 16.7oC to 234oC (3 s.f.) using 725 kJ of energy (3 s.f.).

Assuming an efficiency of 0.342 for the heating process, and that the material does not melt, calculate the specific heat capacity of the material.

m = 4.96 kg
change in T = 217.3 celsius
energy = 725000 J
efficiency = 0.342

Homework Equations


Q = (m)(c)(change in T)

The Attempt at a Solution


c = Q/(m*change in T)

I have calculated Q using .342*725000 = 247950.

I then plugged in these values into the equation and am still getting the wrong answer.

I've also tried dividing 725000/.342 and plugging those values in but it's still incorrect.

What should I be doing in regards to the efficiency?

Thanks!
 
Physics news on Phys.org
  • #2
struggtofunc said:

Homework Statement


A 4.96 kg piece of solid material is heated from 16.7oC to 234oC (3 s.f.) using 725 kJ of energy (3 s.f.).

Assuming an efficiency of 0.342 for the heating process, and that the material does not melt, calculate the specific heat capacity of the material.

m = 4.96 kg
change in T = 217.3 celsius
energy = 725000 J
efficiency = 0.342

Homework Equations


Q = (m)(c)(change in T)

The Attempt at a Solution


c = Q/(m*change in T)

I have calculated Q using .342*725000 = 247950.

I then plugged in these values into the equation and am still getting the wrong answer.

I've also tried dividing 725000/.342 and plugging those values in but it's still incorrect.

What should I be doing in regards to the efficiency?

Thanks!
Please show the details of your calculation of c.
 
  • Like
Likes struggtofunc
  • #3
c = 247950/(4.96*217.3) = 230. 05 J/kg/K
 
  • #4
struggtofunc said:
c = 247950/(4.96*217.3) = 230. 05 J/kg/K
This looks OK, but isn't it usually reported as kJ/kg/K? Also, shouldn't the significant figures be less?
 
  • Like
Likes struggtofunc
  • #5
Hi chestermiller,

Yes, I input the answer as 230 J/kg/K because it was what we were told to present it in. It's still showing as incorrect though.
 
  • #6
struggtofunc said:
Hi chestermiller,

Yes, I input the answer as 230 J/kg/K because it was what we were told to present it in. It's still showing as incorrect though.
I don't know what else to suggest. This answer looks correct.
 
  • #7
No worries. Thank you very much for your help!
 

What is specific heat capacity?

Specific heat capacity is the amount of heat required to raise the temperature of one gram of a substance by one degree Celsius.

How do you calculate specific heat capacity?

Specific heat capacity can be calculated by dividing the amount of heat energy transferred to a substance by its mass and the change in temperature.

What is the unit of specific heat capacity?

The unit of specific heat capacity is joules per gram per degree Celsius (J/g°C) in the SI system.

How does specific heat capacity vary for different substances?

Specific heat capacity varies for different substances due to differences in their atomic and molecular structures. Substances with more complex structures tend to have higher specific heat capacities.

Why is it important to calculate specific heat capacity?

Calculating specific heat capacity is important for understanding how different substances respond to changes in temperature and for predicting the amount of heat energy needed in various processes such as cooking, heating, and cooling.

Similar threads

  • Introductory Physics Homework Help
Replies
8
Views
903
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
968
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
1K
Replies
14
Views
1K
  • Introductory Physics Homework Help
Replies
5
Views
756
  • Introductory Physics Homework Help
Replies
16
Views
4K
  • Introductory Physics Homework Help
Replies
4
Views
1K
  • Introductory Physics Homework Help
Replies
11
Views
1K
Back
Top