Question

A 75-kg copper block initially at 110°C is dropped into an insulated tank that contains 160 L of water at 15°C. Determine the final equilibrium temperature and the total entropy change for this proces

Answer 1

Answer:

T =43.61⁰C

ΔS = 0.84 KJ/K

Explanation:

Given that

Mass of copper block m₁ = 75 kg

Initial temperature of copper block ,T₁ = 110⁰C

Initial temperature of copper block ,water T₂ = 15⁰C

Specific heat for copper ,Cp₁=0.385 J/KkgK

Specific heat for water,Cp₂=4.187 KJ/kgK

Mass pf water ,m₂ = 0.16 x 1000 = 16 kg

Let's take final temperature = T

Heat lost by copper block = Heat gain by water

m₁ Cp₁ (T₁ - T) = m₂Cp₂(T-T₂)

$T=\dfrac{m_1C_{P1}T_1+m_2C_{P2}T_2}{m_1C_{P1}+m_2C_{P2}}$

$T=\dfrac{75\times 0.385\times (273+110)+16\times 4.187\times (273+15)}{75\times 0.385+16\times 4.187}$

T=316.61 K

T =43.61⁰C

The total change in entropy

$\Delta S=m_1C_{P1}\ln\dfrac{T}{T_1}+m_2C_{P2}\ln\dfrac{T}{T_2}$

$\Delta S=75\times 0.385\times\ln\dfrac{316.61}{273+110}+16\times 4.187\times \ln\dfrac{316.61}{273+15}$

ΔS = 0.84 KJ/K