An electric range burner weighing 614 grams is turned off after reaching a temperature of 499.0oC, and is allowed to cool down to 22.2oC.
Calculate the specific heat of the burner if all the heat evolved from the burner is used to heat 565 grams of water from 22.2oC to 80.9oC.
[mass burner x specific heat burner x (Tfinal-Tinitial)] + [mass H2O x speific heat H2O x (Tfinal-Tinitial)] = 0
Substitute and solve for specific heat burner which is the only unknown in the equation..
To calculate the specific heat of the burner, we can use the equation:
Q = m * c * ΔT
where:
Q = heat absorbed or released
m = mass
c = specific heat
ΔT = change in temperature
First, let's calculate the heat released by the burner. The given values are:
mass of the burner (m1) = 614 grams
initial temperature of the burner (T1) = 499.0°C
final temperature of the burner (T2) = 22.2°C
The heat released by the burner (Q1) can be calculated as:
Q1 = m1 * c1 * ΔT1
where c1 is the specific heat of the burner and ΔT1 is the change in temperature.
ΔT1 = T2 - T1 = 22.2°C - 499.0°C = -476.8°C
Q1 = 614 g * c1 * (-476.8°C)
Next, let's calculate the heat absorbed by the water. The given values are:
mass of the water (m2) = 565 grams
initial temperature of the water (T3) = 22.2°C
final temperature of the water (T4) = 80.9°C
The heat absorbed by the water (Q2) can be calculated as:
Q2 = m2 * c2 * ΔT2
where c2 is the specific heat of water and ΔT2 is the change in temperature.
ΔT2 = T4 - T3 = 80.9°C - 22.2°C = 58.7°C
Q2 = 565 g * c2 * 58.7°C
Since all the heat released by the burner is absorbed by the water, Q1 = Q2. Therefore, we can equate the two equations:
614 g * c1 * (-476.8°C) = 565 g * c2 * 58.7°C
Let's solve for c1:
c1 = (565 g * c2 * 58.7°C) / (614 g * (-476.8°C))
c1 ≈ -0.179 g/°C
Therefore, the specific heat of the burner is approximately -0.179 g/°C.
To calculate the specific heat of the burner, we can use the equation:
Q = mcΔT
Where:
Q = heat gained/lost
m = mass
c = specific heat
ΔT = change in temperature
First, let's find the heat lost by the burner when it cools down from 499.0oC to 22.2oC.
ΔT1 = 499.0oC - 22.2oC = 476.8oC
The heat lost by the burner can be calculated using the equation:
Q1 = mcΔT1
Given:
m = 614 grams (mass of the burner)
c = specific heat of the burner (to be calculated)
ΔT1 = 476.8oC
Next, let's find the heat gained by the water when it heats up from 22.2oC to 80.9oC.
ΔT2 = 80.9oC - 22.2oC = 58.7oC
The heat gained by the water can be calculated using the equation:
Q2 = mcΔT2
Given:
m = 565 grams (mass of water)
c = specific heat of water (4.18 J/g.oC) - assuming it is water
ΔT2 = 58.7oC
Since the heat lost by the burner is equal to the heat gained by the water, we have:
Q1 = Q2
Therefore,
mcΔT1 = mcΔT2
We can rearrange the equation to solve for the specific heat of the burner:
c = (mcΔT2) / (mΔT1)
Substituting the given values:
c = (565 g * 4.18 J/g.oC * 58.7oC) / (614 g * 476.8oC)
Now we can calculate the specific heat of the burner.