Question

You mix two sample of water. Sample A is 200g at 20oC. Sample B is 100g at 95oC. Calculate total change in entropy of the process. Assume that specific heat of water is 4190 J·g-1·K-1

Answer 1

$m_1=0.2 kg$ - mass of sample A

$m_2=0.1 kg$ - mass of sample B

$T_1=20^0C+273=293K$ - temperature of sample A

$T_2=95^0C+273=368K$ - temperature of sample B

$c_{water}= 4190 Jkg^{-1}K^{-1}$ //mistake in statements $g \to kg$

Lets agree the temperature of equilibrium $T_3$

According energy conversation law, heating of sample A equl to cooling sample B.

$$c_{water}m_1(T_3-T_1) = c_{water}m_2(T_2-T_3) \implies$$

$$T_3=(m_2T_2+m_1T_1)/(m_1+m_2)$$

Using nubmber $T_3=318K$

$\Delta S_1=c_{water}m_1ln(T_3/T_1)$ - entropy change for sample A

$\Delta S_2=c_{water}m_2ln(T_3/T_2)$ - entropy change for sample B

$\Delta S=\Delta S_1+\Delta S_2$ total entropy change

$\Delta S=7.4582 J/K$

Answer: $7.4582 J/K$