Question

One liter of water at 46◦C is used to make iced tea. How much ice at 0 ◦C must be added to lower the temperature of the tea to 19 ◦C? The specific heat of water is 1 cal/g · ◦ C and latent heat of ice is 79.7 cal/g. Answer in units of g. 

Answer 1

According to the law of conservation of energy;

Heat absorbed = Heat released

1 liter of water = 1000g of water

$m_i(L+ c∆\theta) = m_wc∆\theta\\ m_i(79.7 + 1(19-0)) =1000×1×(46-19)\\ m_i(79.9+19) =27000$

$m_i = \dfrac{27000}{88.9} = 303.71g$

$\therefore$ 303.71g of ice at 0°C must be added to lower the temperature of the tea to 19°C.