Question

The hot water tap of a bath delivers water at 80°c at a rate of 10kg min-1.The cold water tap of the bath delivers water at 20°c at the rate of 20kg min-1. Assuming that both taps are left for 5minutes, calculate the final temperature of the bath water

Answer 1

Solution.

$t_1=80^oC;$

$v_1=10kg/min;$

$t_2=20^oC;$

$v_2=20kg/min;$

$\tau=5min;$

$t-?;$

$m_1=v_1\tau;$

$m_1=10kg/min\sdot5min=50kg;$

$m_2=v_2\tau;$

$m_2=20kg/min\sdot5min=100kg;$

$Q=cm\Delta t;$

$cm_1(t_1-t)=cm_2(t-t_2)\implies t=\dfrac{m_2t_2+m_1t_1}{m_2+m_1};$

$t=\dfrac{100kg\sdot20^oC+50kg\sdot80^oC}{100kg+50kg}=40^oC;$

Answer: $t=40^oC.$