**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>