How much water is needed to cool a block of aluminum to a specific temperature?

  • Thread starter Thread starter PhysicsStuff
  • Start date Start date
  • Tags Tags
    Heat Heat transfer
Click For Summary
SUMMARY

The discussion centers on calculating the mass of water required to cool a 1.85 kg aluminum cube from 150°C to 65°C, using the principle of thermal equilibrium. The correct mass of water needed is 0.845 kg, while the initial incorrect calculation yielded 0.273 kg. Key equations utilized include Q = ΔT * m * C, where ΔT represents the temperature change, m is mass, and C is specific heat. The conversation emphasizes that temperature differences in Celsius and Kelvin are equivalent for this calculation, and Newton's Law of Cooling is not applicable in this context.

PREREQUISITES
  • Understanding of thermal equilibrium concepts
  • Familiarity with specific heat capacity (kJ/kg·K)
  • Basic knowledge of heat transfer equations
  • Ability to perform temperature change calculations
NEXT STEPS
  • Study the principles of thermal equilibrium in heat transfer
  • Learn about specific heat capacities of various materials
  • Explore conservation of energy in thermal systems
  • Investigate the applications of Newton's Law of Cooling in different scenarios
USEFUL FOR

Students in physics or engineering, educators teaching thermodynamics, and anyone interested in heat transfer calculations and thermal equilibrium principles.

PhysicsStuff
Messages
39
Reaction score
0

Homework Statement


What mass of water at 25dC (d = degrees) must be allowed to come to thermal equalibrium with a 1.85kg cube of aluminum initially at 1.5*10*2dC to lower the temperature of the aluminum to 65dC? Assume the water doesn't turn to steam.


Homework Equations


Q = Del(T)m(C), DelT= change in temperature, m = mass, C = specific heat. It doesn't actually give any equation I just assume that it deals with that one.


The Attempt at a Solution


(after conversion to Kelvin)

85K * (1.85kg) * (.91kJ/(kgK)aluminum = 143.0975kJ

143.0975kJ = 125K * m * (4.18kJ/kgK)water

I assumed this was the amount energy involved in the temperature change from aluminum at 423K to 338K, and assumed the energy is conserved since conservation of energy was a topic recently and so assumed the amount of energy that the aluminum cooled by the was amount of energy the water gained. It makes sense at first me at first except I don't understand exactly how to consider the "equilibrium" part of it, I know from calculus there's Newton's law of Cooling and so I don't know why/how that's used with varying temperatures and masses to cool something to an equilibrium instead, and neither do I get exactly how an equilibrium is being considered in this situation. Anyway, it said the answer was .845kg and I got .273kg.
 
Physics news on Phys.org
Why does everyone get their question answered but me :(
 
The final temperature for the water will equal the final temperature for the aluminum, they're in thermal equilibrium.

You listed change in temperature for water as going from initial to aluminums initial, 150. Sub in delta T to be 40 instead of 125.
 
'Thermal equilibrium' means the water and the aluminum have reached the same temperature. As heat is transferred from the aluminum cube to the water, the metal cools while the water heats up.

You aren't being asked how long this process takes, so Newton's Law of Cooling isn't useful for this problem.

Although specific heats are given in units of kJ/K-kg and temperatures are given in degrees C, you don't need to convert temperature from degrees C to degrees K because you are interested in the amount of heat it takes to change the temperature of a substance, and a temperature difference of 1 degree C is the same temperature difference of 1 degree K.

It's not clear why you said the temperature of the water would increase by 125 degrees K. After all, since the water had an initial temperature of 25 C, an increase of 75 C would bring the water to its boiling point of 100 C. That's why your mass calculation failed.
 
I didn't get the exact number but I got pretty close, thanks.
 

Similar threads

  • · Replies 17 ·
Replies
17
Views
6K
  • · Replies 10 ·
Replies
10
Views
3K
  • · Replies 9 ·
Replies
9
Views
3K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
7
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
6
Views
2K
Replies
2
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K