What Is the Equilibrium Temperature of Copper in Water?

[Dx]

Hello,

what will be the equilibrium temperature when a 270 g block of copper at 300 degrees C is placed in a 150 g Al calorimeter cup containing 820 g of water at 12 degree C?

My answer is 20.2 C Is this correct?

I done is like so 270 * (390)*(300-T) = (820)*(900)(T-12) sorry but i forgot where I referenced this formula, I think i am confused with this problem though. It doesn't seem right to me so i thought i would ask?

Thanks!
Dx

 

[quantumdude]

Originally posted by Dx
I done is like so 270 * (390)*(300-T) = (820)*(900)(T-12) sorry but i forgot where I referenced this formula,

The formula is:

Heat lost by copper = Heat gained by water

Where the heat (q) transferred to/from a body is given by:

q=mC(T_hot-T_cold)

m=mass
C=specific heat
T_hot=higher temp
T_cold=lower temp

The problem with your solution is that you need:

Heat lost by copper = Heat gained by water and aluminum[/color]

If you include the aluminum, you will get this right.

 

[M98Ranger]

Hi there,

The equilibrium temperature in this scenario can be calculated using the formula for heat transfer between two objects:

Q1 = Q2

Where Q1 is the heat lost by the copper block and Q2 is the heat gained by the water and calorimeter. We can set up the equation as follows:

(270g)(0.385 J/g°C)(300°C - T) = (820g)(4.18 J/g°C)(T - 12°C)

Simplifying, we get:

103.95(300 - T) = 3427.6(T - 12)

30985.5 - 103.95T = 3427.6T - 41131.2

35399.3 = 3531.55T

T = 10.02°C

So the equilibrium temperature is approximately 10.02°C, not 20.2°C as you calculated. I suggest double-checking your calculations and the formula you used. I hope this helps! Good luck with your studies.