Convert Joules to Calories - 1 cal = 4.184 J

  • Thread starter Thread starter olop4444
  • Start date Start date
  • Tags Tags
    Calories Joules
AI Thread Summary
1 calorie is defined as the energy required to raise the temperature of 1 gram of water by 1 degree Celsius, equating to 4.184 joules. The discussion highlights a misunderstanding in calculating this conversion using equipartition, where the calculated energy was incorrectly determined to be 1.38 joules. The discrepancy arises from the unique properties of water, particularly its hydrogen bonds, which contribute to a higher heat capacity than expected. As water's temperature increases, additional energy is required to break these bonds, leading to the established conversion factor. Understanding these molecular interactions is essential for accurate thermodynamic calculations.
olop4444
Messages
2
Reaction score
0

Homework Statement


1 calorie is the energy needed to raise the temperature of 1 gram of water by 1 degree Celcius, i.e., the specific heat of water is 1 calorie/g-°C. How many joules is 1 calorie?


Homework Equations


ΔU = nαkΔT


The Attempt at a Solution


I attempted to use equipartition to solve this problem.
α of water should be 6*1/2 = 3.
1 gram of water should have 6.02*10^23 * 1/18 = 3.34 * 10^22 molecules
So to increase the temperature of 1 gram of water by 1° C or K, I obtained 3.34*10^22*3*1.38*10^-23*1 = 1.38 J.

However, it is well established that 4.184 J = 1 calorie. What am I doing wrong?
 
Physics news on Phys.org
olop4444 said:

Homework Statement


1 calorie is the energy needed to raise the temperature of 1 gram of water by 1 degree Celcius, i.e., the specific heat of water is 1 calorie/g-°C. How many joules is 1 calorie?

Homework Equations


ΔU = nαkΔT

The Attempt at a Solution


I attempted to use equipartition to solve this problem.
α of water should be 6*1/2 = 3.
1 gram of water should have 6.02*10^23 * 1/18 = 3.34 * 10^22 molecules
So to increase the temperature of 1 gram of water by 1° C or K, I obtained 3.34*10^22*3*1.38*10^-23*1 = 1.38 J.

However, it is well established that 4.184 J = 1 calorie. What am I doing wrong?

Welcome to PF!

Water behaves peculiarly when heated. It is because of its hydrogen bonds. The molecules not only vibrate around their equilibrium positions, but the hydrogen bonds also perform different coupled vibrations. These vibrations are excited more and more between zero and 100 °C. Water has much more degrees of freedom as 6 per molecule. The bonds can also break at higher temperatures, and that also consumes energy. All these result in heat capacity considerably larger than expected.

ehild
 
Interesting. I guess I'll learn more about that later in the course. Thanks for your help!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top