Answer to Question #341309 in Mechanical Engineering for dads

Question

how many kg of hot water at 80C must me added to cold water at 15C in order to make a 1.5kg mixture at 25C? Neglect heat losses

Answer 1

Heat balance says

$$cm_1(t_3-t_1)=cm_2(t_2-t_3)$$

$$m_1+m_2=m$$

$$m_1(25-15)=m_2(80-25)\\ m_1+m_2=1.5$$

Solution of this system of equations

$$m_1=1.27\:{\rm kg}\\ m_2=0.23\:\rm kg$$