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Consider a 1.00 kg sample of solid aluminium (cp,m = 24.35 JK-1mol-1) that is cooled from 300 k to 250 k and calculatern3.2.1 the energy that must be removed as heat. [3]rn3.2.2 the change in entropy of the sample. [3]
Accepted Answer
3.2.1 The energy that must be removed as heat can be calculated using the following equation:
Q = m * Cp,m * ΔT
where:
- Q is the heat energy (in J)
- m is the mass of the sample (in kg)
- Cp,m is the molar specific heat capacity of the sample (in J K^-1 mol^-1)
- ΔT is the change in temperature (in K)
Substituting the given values into the equation, we get:
Q = 1.00 kg * 24.35 J K^-1 mol^-1 * (-50 K)
Q = -1217.5 J
Therefore, 1217.5 J of heat energy must be removed from the sample to cool it from 300 K to 250 K.
3.2.2 The change in entropy of the sample can be calculated using the following equation:
ΔS = - (Q / T)
where:
- ΔS is the change in entropy (in J K^-1)
- Q is the heat energy (in J)
- T is the temperature (in K)
Substituting the given values into the equation, we get:
ΔS = - (-1217.5 J / 283 K)
ΔS = 4.30 J K^-1
Therefore, the change in entropy of the sample is 4.30 J K^-1 when it is cooled from 300 K to 250 K.
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