An 8.6 g sample of a metal is heated to 110.0oC and then placed in a coffee cup calorimeter containing 125 g of water at a temperature of 23.00oC. After the metal cools, the final temperature of the metal and water is 26.83oC. Assuming that no heat is lost to the surroundings or has been absorbed by the calorimeter, calculate the specific heat of the metal. The specific heat of water is 4.18 kJ/L.

heat lost by metal + heat gained by water = 0

[mass metal x specific heat metal x (Tfinal-Tinitial)] + [mass water x specific heat water x (Tfinal-Tinitial)] = 0
Substitute and solve for the one unknown.

To calculate the specific heat of the metal, you can use the principle of heat transfer. The heat gained by the water is equal to the heat lost by the metal.

The heat gained or lost can be calculated using the formula:

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 of the substance (in joules/gram°C)
ΔT is the change in temperature (final temperature - initial temperature)

In this case, the water gains heat, so the formula becomes:

Qwater = mwater * cwater * ΔTwater

And the metal loses heat, so the formula becomes:

Qmetal = mmetal * cmetal * ΔTmetal

Since no heat is lost to the surroundings or absorbed by the calorimeter, Qwater = -Qmetal (negative to indicate heat transfer from metal to water).

We know the following values:
mwater = 125 g
cwater = 4.18 kJ/kg°C (we need to convert it to joules/gram°C)
ΔTwater = Tfwater - Tiwater = 26.83°C - 23.00°C
mmetal = 8.6 g
ΔTmetal = Tfmetal - Timetal = 26.83°C - 110.00°C

Now, let's solve the equation:

Qwater = Qmetal
mwater * cwater * ΔTwater = mmetal * cmetal * ΔTmetal

First, let's convert the specific heat capacity of water from kJ/kg°C to joules/gram°C:

cwaternow = cwater / 1000 = 4.18 kJ/kg°C / 1000 = 0.00418 kJ/g°C

Now, let's calculate Qwater and Qmetal:

Qwater = mwater * cwaternow * ΔTwater
Qmetal = mmetal * cmetal * ΔTmetal

Substituting the given values:

0.00418 kJ/g°C * 125 g * (26.83 - 23.00)°C = cmetal * 8.6 g * (26.83 - 110.00)°C

After solving this equation, you should be able to find the specific heat capacity (cmetal) of the metal.