What is the specific heat capacity of a 250g object that needs 34.125kJ to have its temperature raised by 65.0 ∘ C?
Question
What is the specific heat capacity of a 250g object that needs 34.125kJ to have its temperature raised by 65.0 ∘ C?
Solution
The specific heat capacity (c) of an object can be calculated using the formula:
c = Q / (m * ΔT)
where:
- Q is the heat energy absorbed or released (in joules),
- m is the mass of the object (in kilograms), and
- ΔT is the change in temperature (in degrees Celsius).
Given:
- Q = 34.125 kJ = 34125 J (since 1 kJ = 1000 J),
- m = 250 g = 0.25 kg (since 1 g = 0.001 kg), and
- ΔT = 65.0 °C,
we can substitute these values into the formula to find c:
c = 34125 J / (0.25 kg * 65.0 °C) = 2.1 J/(g°C)
So, the specific heat capacity of the object is 2.1 J/(g°C).
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