Finding the Specific Heat Capacity

[McKeavey]

Homework Statement

The experiment was that we heated a beaker with water in it, and also a type of metal for around 5 minutes so that the metal becomes the same temperature as the water.

We then dropped that heated metal into a stryofoam cup containing 189.10g of water in it.

mass of metal = 64.63 g
initial temperature of the metal = 101.5 degrees celcius
initial temperature of the water in the styrofoam cup = 16 degrees celcius
final temperature of the water in the styrofoam cup (After metal is dropped) = 21 degrees celcius

how do you find the specific heat capacity..?
Thanks!

Homework Equations

No idea..

The Attempt at a Solution

 

[rock.freak667]

Heat given off my the metal = heat gained by the water

you have the final and initial temperatures of the water and the mass and specific heat capacity of the water, so you can get the heat gained.

Do the same with the metal and equate the two.

 

[McKeavey]

Wait so..The Heat capacity of water = 4.3x10^3.
So I figure out the heat energy of water.
and then use that heat energy into
C = mt/Eh to find the heat capacity of metal?

Could you show me an example. I don't fully understand..

 

[rock.freak667]

Wait so..The Heat capacity of water = 4.3x10^3.
So I figure out the heat energy of water.
and then use that heat energy into
C = mt/Eh to find the heat capacity of metal?

Could you show me an example. I don't fully understand..

You know the mass of water, specific heat and temperature difference all of the water.


The heat energy is given by Q=mcΔT.

Q_metal=Q_water

(mcΔT)_metal = (mcΔT)_water

You can calculate the right side easily since you have all the parameters.

On the left side, you can get the temperature difference and you have the mass of the metal.

 

[rwisz]

To find the specific heat capacity, we can use the formula: q = m x c x ΔT, where q is the heat transferred, m is the mass, c is the specific heat capacity, and ΔT is the change in temperature.

First, we need to calculate the heat transferred from the metal to the water. We can do this by using the formula q = mcΔT, where m is the mass of the metal, c is the specific heat capacity of the metal, and ΔT is the change in temperature (final temperature - initial temperature).

We know the mass of the metal (64.63 g), the initial temperature (101.5 degrees Celsius), and the final temperature (21 degrees Celsius). We can assume that the specific heat capacity of the metal is constant, so we can rearrange the equation to solve for c.

c = q / (m x ΔT)

Now, we need to calculate the heat transferred from the metal to the water. We can do this by using the formula q = mcΔT, where m is the mass of the water, c is the specific heat capacity of water (4.18 J/g°C), and ΔT is the change in temperature (final temperature - initial temperature).

We know the mass of the water (189.10 g), the initial temperature (16 degrees Celsius), and the final temperature (21 degrees Celsius). We can now solve for q.

q = (189.10 g) (4.18 J/g°C) (21 - 16) = 1051.358 J

Now, we can plug in the values for q, m, and ΔT into the formula for specific heat capacity.

c = (1051.358 J) / (64.63 g x (21 - 101.5) °C) = 0.22 J/g°C

Therefore, the specific heat capacity of the metal is 0.22 J/g°C.