A piece of iron with a mass of 35 g is heated to 1305oC and then placed in a calorimeter containing 600 g of water. The temperature of the water increases from 25oc to 33oC. The specific heat of water is 4.184 J/g X C. Calculate the specific heat of the iron.
heat lost by Fe + heat gained by H2O = 0
[mass Fe x specific heat Fe x (Tfinal-Tinitial)] + [mass H2O x specific heat water x (Tfinal-Tinitial)]= 0
Substitute and solve for specific heat Fe.
Thanks you Dr BOb
To calculate the specific heat of the iron, we need to use the formula:
Q = mcΔT,
where Q is the heat transferred, m is the mass of the substance, c is the specific heat capacity, and ΔT is the change in temperature.
First, let's calculate the heat transferred to the water:
Q(water) = mcΔT
Q(water) = (600 g)(4.184 J/g°C)(33°C - 25°C)
Q(water) = (600 g)(4.184 J/g°C)(8°C)
Q(water) = 20041.6 J
The heat transferred from the iron is equal to the heat transferred to the water. Therefore:
Q(iron) = Q(water)
Q(iron) = 20041.6 J
Next, we need to calculate the change in temperature for the iron. We know that the initial temperature of the iron is 1305°C, and when it reaches thermal equilibrium with the water, its final temperature is 33°C. Therefore:
ΔT(iron) = 33°C - 1305°C
ΔT(iron) = -1272°C
Now, we can substitute the values we have into the formula to calculate the specific heat of the iron:
Q(iron) = mcΔT(iron)
20041.6 J = (35 g)(c)(-1272°C)
To solve for c, we divide both sides of the equation by (35 g)(-1272°C):
c = 20041.6 J / (35 g)(-1272°C)
Thus, the specific heat of the iron can be calculated using the given information and the formula above.