Answer to Question #194027 in Molecular Physics | Thermodynamics for dudu

Question

In a thermodynamic experiment 25kg of water at 95C is mixed with 35kg of water at 35C, the pressure being taken as constant and the temperature of the surrounding being 15C(Cp of water=4.2kJ/kgK). Calculate total available energy, final temperature after mixing and decrease in available energy due to mixing.

Answer 1

Solution.

$m_1=25kg;$

$T_1=368K;$

$m_2=35kg;$

$T_2=308K;$

$T_0=288K;$

$c_p=4.2KJ/kgK;$

$(A.E.)_{25}=m_1c_p\int_{T_0}^{T_1}(1-\dfrac{T_0}{T})dT=25\sdot4.2\sdot\int_{288}^{368}(1-\dfrac{288}{T})dT=987.49kJ;$

$(A.E.)_{35}=m_2c_p\int_{T_0}^{T_2}(1-\dfrac{T_0}{T})dT=35\sdot4.2\sdot\int_{288}^{308}(1-\dfrac{288}{T})dT=97.59kJ;$

$(A.E.)_{total}=(A.E.)_{25}+(A.E.)_{35}=987.49kJ+97.59kJ=1085kJ;$

$m_1c_p(t_1-t)=m_2c_p(t-t_2) \implies t=\dfrac{m_1t_1+m_2t_2}{m_1+m_2};$

$t=\dfrac{25\sdot95+35\sdot35}{25+35}=60^OC;$

$(A.E.)_{60}=mc_p\int_{T_0}^{T}(1-\dfrac{T_0}{T})dT=35\sdot4.2\sdot\int_{288}^{333}(1-\dfrac{288}{T})dT=803kJ;$

$\Delta(A.E.)=(A.E.)_{total}-(A.E.)_{60}=1085kJ-803kJ=282kJ;$

Answer:$(A.E.)_{total}=1085kJ; t=60^oC;\Delta (A.E.)=282kJ.$