Grade 11 physic problem; calculating heat transfers?

Click For Summary
SUMMARY

The equilibrium temperature calculation involves a 245 g block of copper at 285°C placed in a 145 g aluminum calorimeter cup containing 825 g of water at 12°C. The heat transfer equation used is q=mcΔt, where qlost + qgained = 0. The user correctly sets up the equation but encounters difficulty in simplifying it. The key issue identified is the inconsistency in units for specific heat capacities, which must be standardized for accurate calculations.

PREREQUISITES
  • Understanding of heat transfer principles, specifically q=mcΔt
  • Knowledge of specific heat capacities for copper, aluminum, and water
  • Ability to manipulate algebraic equations and apply the Distributive Law
  • Familiarity with unit conversion and consistency in scientific calculations
NEXT STEPS
  • Review the concept of specific heat capacity and its units
  • Practice solving heat transfer problems using q=mcΔt
  • Learn how to simplify and solve algebraic equations involving multiple variables
  • Explore the implications of thermal equilibrium in different materials
USEFUL FOR

Students studying physics, particularly those focusing on thermodynamics and heat transfer calculations, as well as educators looking for examples of practical applications of these concepts.

raininggently
Messages
2
Reaction score
0

Homework Statement


What will be the equilibrium temperature when a 245-g block of copper at 285° c is placed in a 145 g aluminium calorimeter cup containing 825 g of water at 12°c?


Homework Equations



q=mcΔt

qlost+qgained=0
-qlost=qgained
-m1c1t1=m2c2t2

The Attempt at a Solution


So this is what I did so far:

Know :

Copper Aluminium
- 245 g= 0.245 kg - 145 g= 0.145 kg
- specific heat capacity of copper = 0.385 j kg - ti=12°c
- ti ( Inital temp)= 285°c -teq=?
-teq=?

Water
-825 g=0.825 kg
-ti=12°c
-teq=? ( T equilibrium )

So I know qlost+ qgained= 0

then - qlost=qgained.

In this case the qlost will be the copper.

So this is what I used: ( Cu- copper; al-aluminium; & w- water)

Mcu(Ccu)(Ticu-Teq)=Mal(Cal)(Teq-Ti) + Mw(Cw)(Teq-ti)

(0.245 kg)(0.385)(285°c-Teq)=(0.145 kg)(9.2x10^2)(Teq-12°c)+(0.825kg)(4.18x10^3)(teq-12°c)

Now when I come to this, I have no clue what to do :

(0.094325)(285°c-Teq)=(133.4)(Teq-12°c)+(3448.5)(teq-12°c)

Please let me know what I`m doing wrong, and how i can solve it completely.
 
Physics news on Phys.org
It looks like you're using J/gram/°C for the specific heat of copper, but J/kg/°C for the others. Make sure that you use consistent units for all the values.

As for what to do with the equation once it's written, I'm afraid it needs to be multiplied out and the the Teq terms collected. You might find it easier to do it symbolically, then plug in numbers afterwords.
 
You didn't do anything wrong, you just stopped right before the finish line. You now need to use the Distributive Law of algebra on both sides, collect like terms, and divide. I'll start the left side for you. (0.094325)(285°C-Teq) = 26.88°C - .094325 x Teq = ...
 

Similar threads

  • · Replies 2 ·
Replies
2
Views
9K
  • · Replies 13 ·
Replies
13
Views
2K
Replies
2
Views
5K
  • · Replies 5 ·
Replies
5
Views
8K
  • · Replies 13 ·
Replies
13
Views
2K
  • · Replies 8 ·
Replies
8
Views
5K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
5
Views
3K
  • · Replies 1 ·
Replies
1
Views
4K
Replies
3
Views
2K