Question

Ten kilograms of steam at 100

0C is condensed in 500 kg of water

at 40 0C. What is the resulting temperature? (Ans: 51.8 0C

Answer 1

Solution.

$m_s=10kg;$

$t_s=100^oC;$

$m_w=500kg;$

$t_w=40^oC;$

$r=2.3\sdot10^6J/kg;$

$c=4200 J/{kg^oC};$

$t_f-?;$

$rm_s+cm_s(t_s-t_f)=cm_w(t_f-t_w);$

$rm_s+cm_st_s-cm_st_f=cm_wt_f-cm_wt_w;$

$rm_s+cm_st_s+cm_wt_w=cm_wt_f+cm_st_f;$

$t_f(cm_w+cm_s)=rm_s+cm_st_s+cm_wt_w;$

$t_f=\dfrac{rm_s+cm_st_s+cm_wt_w}{cm_w+cm_s};$

$t_f=\dfrac{2.3\sdot10^6\sdot10+4200\sdot10\sdot100+4200\sdot500\sdot40}{4200(10+500)}=51.9^oC;$

Answer: $t_f=51.9^oC.$