Question

Calculate the final equilibrium temperature when 150 grams of 23 degrees Celsius water are mixed with 50 grams of ice at 0 degrees Celsius. The specific heat of water is 4186J/Kg C and the heat of fusion for water is 334 Kj/kg

Answer 1

$Q_1=Q_2,$

$Q_1=cm_1(t_1-t),$

$Q_2=m_2(\lambda+c(t-t_2)),$

$t=\frac{\lambda m_2-cm_1t_1}{c(m_1+m_2)}=2.7°C.$