Answer :
Final answer:
The entropy change of the surroundings in J/mol∙K when 30 kJ of heat is released by the system at 27°C is calculated as -100 J/mol∙K, although this option is not given in the provided choices. This calculation involves converting heat and temperature units and applying the formula ∆S = q/T.
Explanation:
In this problem, we are asked to calculate the entropy change of the surroundings, given that 30 kJ of heat is released by the system at 27°C. The formula to calculate the entropy change (∆S) is ∆S = q/T, where q is the heat transferred and T is the absolute temperature.
First, we convert the heat from kJ to J, because entropy is typically expressed in terms of J. Thus, 30 kJ = 30,000 J. The heat is released, so it is negative, -30,000 J.
Next, we convert the temperature from °C to Kelvin (K) by adding 273.15 to the Celsius temperature to maintain the correct scale. So, 27°C = 300.15 K.
Last, we substitute these values into the formula: ∆S = q/T = (-30,000 J)/(300.15 K) = -100 J/K. However, given the choices, we need to convert this to J/mol∙K. Assuming 1 mol, the answer will be -100 J/mol∙K, which is not in the options given. Please recheck the conditions provided.
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