Question

a) Calculate the change in entropy when 50 g. of water at 800 C is poured into 100 g. of water at 100 C in an insulated vessel.

I'm not sure what the answer is please try your best.

Answer 1

Ans:

We know that during mixing

“heat lost by the hot water (q1) = heat gained by the cold water (q2)”

So, q1 = -q2

Also, we know that, q = m x Cp x ∆T

where, m = mass of water,

Cp = specific heat of water = 4179.6 J·kg⁻¹·K⁻¹ (data collected)

∆T = change in temperature = (Tf-Ti).

Let ‘T’ be the equilibrium temperature after mixing. So, for both cold and hot water, final temperature, Tf will be T.

q1 = -q2

(m x Cp x ∆T)_hot = -( m x Cp x ∆T)_cold

(50 x Cp x (T-800)) = -(100 x Cp x (T - 100))

T = 333.33 ^oC

Since entropy is a state-function, we can see the whole process in two steps, the

cooling of the hot water and the heating of the cold water:

∆S = ∆S_hot + ∆S_cold

∆S_hot = m.Cp.ln(Tf/Ti) = 50 x 10^-3Kg x 4179.6 J/kg.K x ln((333.33+273)/(800+273)) = -119.2836 J

∆S_cold = m.Cp.ln(Tf/Ti) = 100 x 10^-3Kg x 4179.6 J/kg.K x ln ((333.33+273)/(100 +273)) = 203.0641 J

∆S = ∆S_hot + ∆S_cold = -119.2836 + 203.0641 = 83.7805 J