10.0 grams of octane are burned in a bomb calorimeter containing 2.00 × 102 g H2O. If the water temperature increases from 25.00°C to 37.00°C, how much energy was released by the system? [SHw = 1.00 cal/(g•°C)]
Question
10.0 grams of octane are burned in a bomb calorimeter containing 2.00 × 102 g H2O. If the water temperature increases from 25.00°C to 37.00°C, how much energy was released by the system? [SHw = 1.00 cal/(g•°C)]
Solution
To solve this problem, we need to use the formula for heat transfer:
q = m * c * ΔT
where:
- q is the heat energy,
- m is the mass of the substance (in this case, water),
- c is the specific heat capacity of the substance (for water, it's given as 1.00 cal/(g•°C)),
- ΔT is the change in temperature.
Given: m = 2.00 × 10^2 g (mass of water) c = 1.00 cal/(g•°C) (specific heat capacity of water) ΔT = 37.00°C - 25.00°C = 12.00°C (change in temperature)
Substitute these values into the formula:
q = (2.00 × 10^2 g) * (1.00 cal/(g•°C)) * (12.00°C) q = 2400 cal
So, the energy released by the system is 2400 calories.
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