Grade 11 physic problem; calculating heat transfers?

In summary: I'll leave it to you to finish the right side and combine them all on one side.In summary, 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 can be found by using the formula q=mcΔt and considering the specific heat capacities of each material. After solving the equation, the equilibrium temperature is approximately 40.4°C.
  • #1
raininggently
2
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.
 
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  • #2
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.
 
  • #3
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 = ...
 

1. How do you calculate heat transfer?

To calculate heat transfer, you need to know the mass of the substance, the specific heat capacity of the substance, and the change in temperature. The formula for heat transfer is Q = mcΔT, where Q is the heat transfer, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

2. What is the difference between conduction, convection, and radiation?

Conduction is the transfer of heat through direct contact between two substances. Convection is the transfer of heat through the movement of fluids or gases. Radiation is the transfer of heat through electromagnetic waves.

3. How do you convert between Celsius and Fahrenheit?

To convert from Celsius to Fahrenheit, use the formula F = (C x 9/5) + 32, where F is the temperature in Fahrenheit and C is the temperature in Celsius. To convert from Fahrenheit to Celsius, use the formula C = (F - 32) x 5/9.

4. What is the specific heat capacity of water?

The specific heat capacity of water is 4.184 J/g°C. This means that it takes 4.184 Joules of energy to raise the temperature of 1 gram of water by 1 degree Celsius.

5. How does the type of material affect heat transfer?

The type of material affects heat transfer because different substances have different specific heat capacities and conduct heat at different rates. Materials with high specific heat capacities require more energy to change temperature, while materials with high conductivity transfer heat more easily. Additionally, the state of the material (solid, liquid, gas) can also affect heat transfer. For example, water has a higher specific heat capacity than most other substances, making it better at regulating temperature and storing heat.

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