A 30-g iron bar at 100degreeC is placed in 200 g of water at 20degreeC. If the specific heat capacity of iron is 0.11 cal/gdegreeC, to what final temperature will the iron bar cool?
q = mcT
Find the q for water (change in temp should be 100 - 20)
q = 200(c)(80)
Then solve for the final temp of the iron using -q.
To find the final temperature to which the iron bar will cool, we need to use the principle of conservation of energy. The heat lost by the iron bar will be equal to the heat gained by the water.
First, let's calculate the initial heat of the iron bar using the formula:
Q = mcΔT
Where:
Q = Heat gained or lost (in calories)
m = Mass of the iron bar (in grams)
c = Specific heat capacity of iron (in cal/g°C)
ΔT = Change in temperature (final temperature - initial temperature)
For the iron bar:
m = 30 g
c = 0.11 cal/g°C
Initial temperature of the iron bar (Ti) = 100°C
Q_iron = mcΔT_iron
Next, let's calculate the heat gained by the water. We can use the formula:
Q = mcΔT
For the water:
m = 200 g
c = 1 cal/g°C (specific heat capacity of water)
Initial temperature of the water (Ti) = 20°C
Final temperature of the water (Tf) = Final temperature of the iron bar
Q_water = mcΔT_water
According to the principle of conservation of energy, the heat lost by the iron bar is equal to the heat gained by the water. Therefore, we can equate these two equations:
mcΔT_iron = mcΔT_water
Substituting the given values:
(30 g) * (0.11 cal/g°C) * (100°C - Tf) = (200 g) * (1 cal/g°C) * (Tf - 20°C)
Now, we can solve this equation to find the value of Tf.
First, let's multiply out the equation:
3.3 (100 - Tf) = 200 (Tf - 20)
330 - 3.3Tf = 200Tf - 4000
Combine like terms:
3.3Tf + 200Tf = 4000 + 330
203.3Tf = 4330
Divide both sides of the equation by 203.3:
Tf ≈ 21.3°C
Therefore, the final temperature to which the iron bar will cool is approximately 21.3°C.