A 47.5g sample of metal is heated to 300.0ºC and placed in a calorimeter containing 75.0g water at 22.0ºC. The final temperature of the system is 40.0ºC. What is the specific heat of the metal? (the specific heat of water is 4.18 J/g ºC)

Thanks for your time helping!

The heat lost by the metal is the heat gained by the water

heat change is mass * delta T * spec heat
Let the spec heat of the metal be z.

47.5(300-40)z = 75(40-22)(4.18)
z = 0.457

Looks like the metal is most likely iron.

To find the specific heat of the metal, we can use the principle of energy conservation. The heat gained by the metal will be equal to the heat lost by the water. The formula used to calculate the heat gained or lost is:

q = m * c * ΔT

Where:
q = heat gained or lost (in joules)
m = mass of the substance (in grams)
c = specific heat capacity of the substance (in J/g ºC)
ΔT = change in temperature (in ºC)

First, let's calculate the heat gained by the metal:

q(1) = m(1) * c(1) * ΔT(1)
q(1) = 47.5g * c(1) * (40.0ºC - 300.0ºC)
The mass of the metal is 47.5g, the change in temperature is (40.0ºC - 300.0ºC = -260.0ºC)

Next, let's calculate the heat lost by the water:

q(2) = m(2) * c(2) * ΔT(2)
q(2) = 75.0g * 4.18 J/g ºC * (40.0ºC - 22.0ºC)
The mass of the water is 75.0g, the specific heat of water is 4.18 J/g ºC, and the change in temperature is (40.0ºC - 22.0ºC = 18.0ºC)

According to the principle of energy conservation, the heat gained by the metal (q(1)) will be equal to the heat lost by the water (q(2)):

q(1) = q(2)

Plugging in the values obtained:

47.5g * c(1) * (-260.0ºC) = 75.0g * 4.18 J/g ºC * 18.0ºC

Now we can solve for c(1), the specific heat capacity of the metal.

c(1) = (75.0g * 4.18 J/g ºC * 18.0ºC) / (-260.0g * ºC)

Evaluating this expression will give us the specific heat capacity of the metal.