Question

A cup containing 200 g of hot water is taken off the stove placed on the kitchen table. Initially the water is at 75Degree C. but it cools down spontaneously and comes to equilibrium with the room at 21 C. The process takes places at constant pressure. What are Delta H, Q, and Delta S for the water? What is Delta S_surrounding? What is Delta S_universe? Take Cp = 4. 184 J/(mole. K) for water and assume that this is approximately constant over temperature range of interest.

Answer 1

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

Step-by-step explanation:

Given:

Cp = 4. 184 J/(mole. K)

T₁ = 75 ⁰C

T₂ = 21 ⁰C

Mass of water = 200 g = 0.2 kg

Since,

`\Delta H=m* C* (T_f-T_i)`

ΔH = 0.2*4.184*(21-75) kJ

ΔH = -45.1872 kJ , where negative sign signifies heat loss.

Since the process is at constant pressure

Q = ΔH = -45.1872 kJ , where negative sign signifies heat loss.

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:

S = m×Cp×ln(T₂ /T₁)

S = 0.2*4.184*ln(294.15/348.15)

S system = -0.141 kJ/K

The heat gain by surroundings

dQ = -Qreaction = 45.1872 kJ

The entropy change of surroundings is

S surr = dQ/T₂ = 45.1872/294 .15

S surr = 0.1536 kJ/K

The entropy of universe is the sum total of the entropy of the system and the surroundings and thus,

S universe = Ssys + Ssurr

S universe = -0.141 kJ/K + 0.1536 kJ/K = 0.0126 kJ/K