Question

2. A 50 g Aluminium disk at 300 oC is placed in 200 cm3 of ethyl alcohol at 10 oC, then removed quickly. The Aluminium temperature is found to have dropped to 120 oC. Calculate the new temperature of the ethyl alcohol.

Answer 1

Solution.

$m_a=0.05kg;$

$t_{a1}=300^oC;$

$t_{a2}=120^oC;$

$c_a=920J/kg^oC;$

$V_{e.a}=200cm^3;$

$t_{e.a}=10^oC;$

$\rho_{e.a}=0.8g/cm^3;$

$m_{e.a}=V{e.a}\sdot \rho_{e.a};$

$c_{e.a}=2500J/kg^oC;$

$m_{e.a}=200\sdot0.8=160g=0.16kg;$

$Q_a=c_am_a(t_{a1}-t_{a_2});$

$Q_a=920\sdot0.05\sdot180=8280J;$

$Q_{e.a}=c_{e.a}\sdot m_{e.a} \sdot(t_f-t_{e.a});$

$Q_{e.a}=Q_a;$

$t_f=\dfrac{Q_{e.a}}{m_{e.a}c_{e.a}}+t_{e.a};$

$t_f=\dfrac{8280}{0.16\sdot2500}+10=30.7^oC;$

Answer:$t_f=30.7^oC.$