Note the correct spelling of celsius.
q = mass H2O x specific heat H2O x delta T.
q = mass H2O x specific heat H2O x delta T.
Q = m × C × ΔT
Where:
Q = heat transferred (energy released)
m = mass of water
C = specific heat capacity of water
ΔT = change in temperature
First, let's find the values needed to plug into the formula:
1. Mass of water (m):
The problem states that there are 2.00x10^2 g of water in the calorimeter. Therefore, m = 2.00x10^2 g.
2. Specific heat capacity of water (C):
The specific heat capacity of water is approximately 1.00 cal/g°C. Therefore, C = 1.00 cal/g°C.
3. Change in temperature (ΔT):
The water temperature increased by 6.00°C. Therefore, ΔT = 6.00°C.
Now, substitute the values into the formula:
Q = (2.00x10^2 g) × (1.00 cal/g°C) × (6.00°C)
Calculate the value:
Q = 1200 cal
Therefore, the amount of energy released during the combustion of octane is 1200 calories.
q = mcΔT
Where:
- q is the amount of energy released (in calories)
- m is the mass of water (in grams)
- c is the specific heat capacity of water (1 cal/g°C)
- ΔT is the change in temperature (in °C)
Given:
- Mass of water (m) = 2.00x10^2 g
- Change in temperature (ΔT) = 6.00 °C
To calculate the amount of energy released, we need to calculate the specific heat capacity of octane (c_octane) first.
The specific heat capacity of water (c_water) is 1 cal/g°C.
Assuming that the octane completely burns, we can use the law of conservation of energy:
q_octane = -q_water
Let's calculate the amount of energy released by the water first:
q_water = m x c_water x ΔT
= (2.00x10^2 g) x (1 cal/g°C) x (6.00 °C)
= 1200 cal
Since the water releases the same amount of energy as the octane absorbs, the amount of energy released by the octane is 1200 cal.
Therefore, the amount of energy released if the water temperature increases by 6.00 °C is 1200 calories.