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)
[tex]T=\dfrac{m_1C_{P1}T_1+m_2C_{P2}T_2}{m_1C_{P1}+m_2C_{P2}}[/tex]
[tex]T=\dfrac{75\times 0.385\times (273+110)+16\times 4.187\times (273+15)}{75\times 0.385+16\times 4.187}[/tex]
T=316.61 K
T =43.61⁰C
The total change in entropy
[tex]\Delta S=m_1C_{P1}\ln\dfrac{T}{T_1}+m_2C_{P2}\ln\dfrac{T}{T_2}[/tex]
[tex]\Delta S=75\times 0.385\times\ln\dfrac{316.61}{273+110}+16\times 4.187\times \ln\dfrac{316.61}{273+15}[/tex]
ΔS = 0.84 KJ/K
