Answer:
ΔH = -45.1872 kJ , where negative sign signifies heat loss.
Q = ΔH = -45.1872 kJ , where negative sign signifies heat loss.
S system = -0.141 kJ/K
S surroundings = 0.1536 kJ/K
S universe = -0.141 kJ/K + 0.1536 kJ/K = 0.0126 kJ/K
Explanation:
Given:
Cp = 4. 184 J/(mole. K)
T₁ = 75 ⁰C
T₂ = 21 ⁰C
Mass of water = 200 g = 0.2 kg
Since,
ΔH = 0.2*4.184*(21-75) kJ
<u> ΔH = -45.1872 kJ , where negative sign signifies heat loss.</u>
Since the process is at constant pressure
<u> Q = ΔH = -45.1872 kJ , where negative sign signifies heat loss.</u>
The conversion of T( °C) to T(K) is shown below:
T(K) = T( °C) + 273.15
T₁ = 75 ⁰C = 348 .15 K
T₂ = 21 ⁰C = 294.15 K
The entropy of the water is given by:
<u> S = m×Cp×ln(T₂ /T₁) </u>
S = 0.2*4.184*ln(294.15/348.15)
<u> S system = -0.141 kJ/K</u>
The heat gain by surroundings
<u> dQ = -Qreaction = 45.1872 kJ </u>
The entropy change of surroundings is
S surr = dQ/T₂ = 45.1872/294 .15
<u> S surr = 0.1536 kJ/K </u>
The entropy of universe is the sum total of the entropy of the system and the surroundings and thus,
S universe = Ssys + Ssurr
<u> S universe = -0.141 kJ/K + 0.1536 kJ/K = 0.0126 kJ/K</u>