A 46.2 gram sample of copper is heated to 95.4 degrees celcius and then place into a calorimeter containing 75.0 grams of water at 19.6 degrees celcius. The final temp of the water and copper is 21.8 degrees celcius. Calculate the specific heat capacity of copper, assuming all heat lost by the copper is gained by the water.

Answer in J/gram degree celcius

how do i set this problem up? I already sorted the information.

[mass Cu x specific heat Cu x (Tfinal-Tinitial)] + [mass H2O x specific heat H2O x (Tfinal-Tinitial)] = 0

Solve for specific heat Cu. It's easier to put in all the numbers first into the above, do the math, then solve for specific heat Cu last. Post your work if you get stuck. For your information, note the correct spelling of Celsius.

Well, it sounds like you've got everything sorted, which is a good start! Now, let's dive into setting up the problem.

First, we need to determine the heat gained by the water. We can use the formula:

Qwater = mcΔT

Where:
- Qwater is the heat gained by the water
- m is the mass of the water
- c is the specific heat capacity of water
- ΔT is the change in temperature (final temperature minus initial temperature)

Next, we need to determine the heat lost by the copper. Again, we can use the same formula:

Qcopper = mcΔT

Where:
- Qcopper is the heat lost by the copper
- m is the mass of the copper
- c is the specific heat capacity of copper
- ΔT is the change in temperature (final temperature minus initial temperature)

Since all the heat lost by the copper is gained by the water, we can equate Qwater to -Qcopper (the negative sign indicates heat lost by the copper):

mcΔT = -mcΔT

Now, we can plug in the values we know:

ma(ΔT)a = -mb(ΔT)b

Where:
- ma is the mass of the copper (46.2 grams)
- ΔTa is the change in temperature for the copper (final temperature minus initial temperature for copper)
- mb is the mass of the water (75.0 grams)
- ΔTb is the change in temperature for the water (final temperature minus initial temperature for water)

Simplifying the equation further:

ma(ΔT)a = -mb(ΔT)b
46.2g(ΔTa) = -75.0g(ΔTb)

Now, we know the final temperature of the water and copper (21.8 degrees Celsius) and the initial temperature of the water (19.6 degrees Celsius). So, we can substitute these values into the equation:

46.2g(21.8°C - Tf) = -75.0g(21.8°C - 19.6°C)

Solving this equation will give you the specific heat capacity of copper (c). To get it into J/gram degree Celsius, you'll need to make sure to convert the final Celsius temperature to Kelvin.

I hope this helps you set up the problem! If you have any other questions or need more clarification, feel free to ask!

To solve this problem, you need to use the principle of conservation of energy, which states that the heat lost by one object is equal to the heat gained by another object. In this case, all the heat lost by the copper will be gained by the water.

The formula we will use is:

Q = m*C*ΔT

Where:
Q is the heat gained or lost (in joules)
m is the mass of the substance (in grams)
C is the specific heat capacity (in J/g·°C)
ΔT is the change in temperature (final temperature - initial temperature) (in °C)

For the copper:
Initial temperature: T1 = 95.4 °C
Final temperature: T2 = 21.8 °C
Mass: m1 = 46.2 g

For the water:
Initial temperature: T1 = 19.6 °C
Final temperature: T2 = 21.8 °C
Mass: m2 = 75.0 g

Since the heat lost by the copper is equal to the heat gained by the water, we can set up the equation as follows:

(m1*Cc*(T2 - T1)) = (m2*Cw*(T2 - T1))

where:
Cc is the specific heat capacity of copper
Cw is the specific heat capacity of water

Simplifying the equation:

Cc = (m2*Cw*(T2 - T1)) / (m1*(T2 - T1))

Plugging in the given values:

Cc = (75.0 g * 4.18 J/g·°C * (21.8 °C - 19.6 °C)) / (46.2 g * (21.8 °C - 95.4 °C))

Simplifying further:

Cc = (75.0 g * 4.18 J/g·°C * 2.2 °C) / (-46.2 g * 73.6 °C)

Finally:

Cc ≈ -0.014 J/g·°C

Keep in mind that the specific heat capacity should always be positive. Since the result is negative, there might be an error in the calculations. Double-check the values and ensure that the temperatures and masses are correct.

To set up this problem, you need to use the principle of heat transfer, which states that the heat lost by the copper is equal to the heat gained by the water in the calorimeter.

First, let's list the given information:
Mass of copper (mC) = 46.2 grams
Initial temperature of copper (TCi) = 95.4 degrees Celsius
Final temperature of copper and water (TCf) = 21.8 degrees Celsius

Mass of water (mW) = 75.0 grams
Initial temperature of water (TWi) = 19.6 degrees Celsius

Next, calculate the change in temperature for both copper and water:

Change in temperature (∆TC) = TCf - TCi = 21.8°C - 95.4°C = -73.6 degrees Celsius

Change in temperature (∆TW) = TCf - TWi = 21.8°C - 19.6°C = 2.2 degrees Celsius

Since all heat lost by the copper is gained by the water, we can set up the equation:

Heat lost by copper = Heat gained by water

The equation for heat transfer is given by:

q = mc∆T

Where:
q is the heat energy transferred
m is the mass of the substance
c is the specific heat capacity of the substance
∆T is the change in temperature

Substituting the values into the equation:
(mC * cC * ∆TC) = (mW * cW * ∆TW)

We want to solve for the unknown specific heat capacity of copper (cC).

Rearranging the equation:
cC = (mW * cW * ∆TW) / (mC * ∆TC)

Plugging in the values given:
cC = (75.0g * cW * 2.2°C) / (46.2g * -73.6°C)

Note that the specific heat capacity of water (cW) is generally considered to be 4.184 J/g°C.

Thus, the formula to solve for the specific heat capacity of copper (cC) is:
cC = (4.184 J/g°C) * (75.0g * 2.2°C) / (46.2g * -73.6°C)

Performing the calculations will give you the specific heat capacity of copper in J/g°C.

heat = (sp. heat)(grams)(temp. change)

or q = (C)(m)(Tf-Ti)
Heat released by Cu = Heat absorbed by H2O
(C)(46.2g)(95.4C-21.8C)= (4.18J/g.C)(75.0g)(21.8C-19.6C)
Solve for C to get the specific heat of copper.