Specific heat of a piece of metal dropped in water

In summary, the specific heat of the metal in a metal container with 14.3 kg of water is being calculated by using the equation Q = c * m * change in T. The final temperature of the system is 17.1°C and the initial temperature of the metal is 190.0°C. By solving for Cmetal, the specific heat of the metal can be found.
  • #1
cyclonefb3
7
0

Homework Statement



A metal container, which has a mass of 8.0 kg contains 14.3 kg of water. A 2.9-kg piece of the same metal, initially at a temperature of 190.0°C, is dropped into the water. The container and the water initially have a temperature of 15.1°C and the final temperature of the entire system is 17.1°C. Calculate the specific heat of the metal.

Homework Equations



Q = c* m *change in T

The Attempt at a Solution



The heat of the container + the heat of the water = the heat of the piece of metal

using algebra I got the equation:

cwater*Mwater*change in Twater / [Mpiece*change in Tpiece - Mcontainer*change in Tcontainer]

I am having trouble figuring out the temperatures for the equation. Please help!
 
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  • #2
Nearly right, The change in T for the container and the water is the same.

So the equation is
Cwater * Mwater * dT water + Cmetal * MContainer * dTwater = Cmetal * Mmetal *dTmetal

The only unknown is Cmetal.
 
  • #3
Hi cyclonefb3,

There's no equality sign in your eqn.

Let t be the final temp. Then,

(14.3*Cw+8*Cm)(17.1-15.1) = (2.9*Cm)(190-17.1).

If Cwater is known, you can find Cm.
 
  • #4
I think I have the dTpiece wrong this is what I'm entering:

41.86(14.3)(17.1-15.1) / [2.9*(190-15.1)] - [8*(17.1-15.1)]

I'm not sure where to go from here...
 
  • #5
It would help if your equations had = signs!

Cwater * Mwater * dT water + Cmetal * MContainer * dTwater = Cmetal * Mmetal *dTmetal
4200 * 14.3 (17.1-15.1) + C * 8 (17.1-15.1) = C * 2.9 (190-17.1)
120120 = C (2.9*172.9 - 8*2)

Hint C for metals is generally a few hundred J/kg/k
 
Last edited:
  • #6
thanks, it was just he program being really picky with digits.
 

1. What is specific heat and why is it important for metal and water?

Specific heat is the amount of heat energy required to raise the temperature of a substance by 1 degree Celsius. It is important for metal and water because it determines how they will react to changes in temperature, such as when a piece of metal is dropped in water.

2. How does the specific heat of metal affect its temperature when dropped in water?

The specific heat of metal determines how much heat energy it can absorb or release. When a metal with a high specific heat is dropped in water, it will absorb more heat energy before its temperature increases, resulting in a slower temperature change compared to a metal with a lower specific heat.

3. Does the size or shape of the metal affect its specific heat when dropped in water?

No, the size or shape of the metal does not affect its specific heat. The specific heat is a property of the material itself and is not dependent on its size or shape.

4. How does the specific heat of water affect the temperature change when a metal is dropped in it?

The specific heat of water is much higher than that of most metals. This means that water can absorb more heat energy before its temperature changes significantly. Therefore, when a metal is dropped in water, it will cause a greater temperature change in the water compared to the metal.

5. Is specific heat the only factor that affects the temperature change when a metal is dropped in water?

No, other factors such as the initial temperature of the metal and water, the mass of the metal and water, and the rate at which heat is transferred can also affect the temperature change when a metal is dropped in water.

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