Question

A 2.64-kg copper part, initially at 400 K, is plunged into a tank containing 4 kg of liquid water, initially at 300 K. The copper part and water can be modeled as incompressible with specific heats 0.385 kJ/kg ? K and 4.2 kJ/kg ???? K, respectively. For the copper part and water as the system, determine (a) the final equilibrium temperature, in K, and (b) the amount of entropy produced within the tank, in kJ/K. Ignore heat transfer between the system and its surrounding

Answer 1

Answer:

a) $T_f=305.7049\ K$

b) $\Delta S=313.51\ J.K^{-1}$

Explanation:

Given:

• mass of copper, $m_c=2.64\ kg$

• initial temperature of copper, $T_{ic}=400\ K$

• specific heat capacity of copper, $c_c=385\ J.kg^{-1}.K^{-1}$

• mass of water, $m_w=4\ kg$

• initial temperature of water, $T_{iw}=300\ K$

• specific heat capacity of water, $c_w=4200\ J.kg^{-1}.K^{-1}$

a)

∵No heat is lost in the environment and the heat is transferred only between the two bodies:

Heat rejected by the copper = heat absorbed by the water

$2.64\times 385\times (400-T_f)= 4\times 4200\times (T_f-300)$

$T_f=305.7049\ K$

b)

Now the amount of heat transfer:

$Q=m_c.c_c.(T_{ic}-T_{f})$

$Q=2.64\times 385\times (400-305.7049)$

$Q=95841.5841\ J$

∴Entropy change

$\Delta S=\frac{dQ}{T}$

$\Delta S=\frac{95841.5841}{305.7049}$

$\Delta S=313.51\ J.K^{-1}$