If the water had lost 12000 joules and the calorimeter gained 1300 joules as the temperature of both rock and calorimeter increased from 22°C to 73°C and the mass of rock in the calorimeter was 189 grams, what is the specific heat, c, of the rock material?

If the water had lost 12000 joules and the calorimeter gained 1300 joules, then the rock gained the difference in energy.

c = ∆E/m∆T = (12000-1300)/(73-22) [J/g·K]

To find the specific heat, c, of the rock material, we can use the formula:

Q = mcΔT

Where:
Q is the heat gained or lost
m is the mass of the substance
c is the specific heat of the substance
ΔT is the change in temperature

In this case, we are given:
Q water = -12000 joules (negative because the water lost heat)
Q calorimeter = 1300 joules (positive because the calorimeter gained heat)
m rock = 189 grams
ΔT = (73°C - 22°C) = 51°C

First, let's calculate the amount of heat gained or lost by the water:

Q water = mcΔT
-12000 J = (m water) (c water) (ΔT)
-12000 J = (m water) (4.18 J/g°C) (51°C) **Note: The specific heat of water is 4.18 J/g°C

We can rearrange this equation to solve for the mass of water (m water):

m water = -12000 J / [(4.18 J/g°C) (51°C)]

Next, let's calculate the specific heat of the rock material:

Q calorimeter = mcΔT
1300 J = (m rock) (c rock) (ΔT)

We can rearrange this equation to solve for the specific heat of the rock material (c rock):

c rock = 1300 J / [(m rock) (ΔT)]

Now we substitute the given values into the formulas:

m water = -12000 J / [(4.18 J/g°C) (51°C)]
c rock = 1300 J / [(189 g) (51°C)]

After evaluating these calculations, you will have the specific heat, c, of the rock material.