If 4.00g of iron metal are reacted in a calorimeter with a heat capacity of 500J/ degree Celsius, what would be the final temperature of the calorimeter if its initial temperature were 23.0 degree Celsius?

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To find the final temperature of the calorimeter, let's determine the heat gained or lost in the reaction. We can use the formula:

q = m × C × ΔT

Where:
q is the heat gained or lost
m is the mass of the substance (in this case, the iron metal)
C is the specific heat capacity of the substance (in this case, the calorimeter)
ΔT is the change in temperature (final temperature - initial temperature)

Given:
Mass of iron metal (m) = 4.00g
Specific heat capacity of calorimeter (C) = 500J/°C
Initial temperature of calorimeter = 23.0°C

First, we need to calculate the heat gained or lost by the reaction. Since the calorimeter absorbs or releases heat, the equation becomes:

q = - q(calorimeter)

Next, we need to calculate q(calorimeter) using the formula:

q(calorimeter) = m × C × ΔT(calorimeter)

Since the calorimeter is at the initial temperature, the change in temperature (ΔT(calorimeter)) will be the difference between the final temperature (Tf) and the initial temperature (Ti = 23.0°C).

So, the equation becomes:

q(calorimeter) = m × C × (Tf - Ti)

Rearranging the equation, we get:

(Tf - Ti) = q(calorimeter) / (m × C)

Substituting the given values into the equation:

(Tf - 23.0°C) = q(calorimeter) / (4.00g × 500J/°C)

Now, we need to determine q(calorimeter). Since the reaction is conducted in the calorimeter, the heat gained or lost by the iron (q) is equal in magnitude but opposite in sign to the heat gained or lost by the calorimeter (q(calorimeter)).

Therefore:

q = -q(calorimeter)

Let's calculate q using the equation:

q = m × C × ΔT

The mass of the iron (m) is given as 4.00g, and the specific heat capacity of iron (C) is approximately 0.45 J/g°C.

q = 4.00g × 0.45J/g°C × ΔT
q = 1.80J/°C × ΔT

Since q is equal in magnitude but opposite in sign to q(calorimeter), the equation becomes:

-1.80J/°C × ΔT = q(calorimeter)

Now, we can substitute the value of q(calorimeter) into the previous equation:

(Tf - 23.0°C) = -1.80J/°C × ΔT / (4.00g × 500J/°C)

Now, we can calculate ΔT:

(Tf - 23.0°C) = -0.009J/°C / (2.00J/°C)
(Tf - 23.0°C) = -0.0045
Tf - 23.0°C = -0.0045

Finally, we can isolate Tf to find the final temperature of the calorimeter:

Tf = -0.0045 + 23.0°C
Tf = 22.9965°C

Rounding this value to the appropriate number of significant figures gives us:

Tf = 23.0°C

Therefore, the final temperature of the calorimeter would be 23.0°C.