Question

An insulated beaker with negligible mass contains a mass of 0.350 kg of water at a temperature of 69.4 ∘C.

How many kilograms of ice at a temperature of − 14.7 ∘C must be dropped in the water to make the final temperature of the system 29.3 ∘C?

Take the specific heat for water to be 4190 J/(kg⋅K) , the specific heat for ice to be 2100 J/(kg⋅K) , and the heat of fusion for water to be 334 kJ/kg .

Answer 1

When ice at -14.7^oC is dropped into water and brought to a temperature of 29.3^oC, it undergoes 3 state changes, i.e. releases heat in 3 forms. -

1) Temperature of ice rising from -14.7^oC to 0^oC = Q = m_ice x c_ice x (T_f-T_i) = m_icex(2100)x(0-(-14.7)

= 30,870 J/kg x m_ice

2) Latent heat of fusion released by ice = Q_f = m_ice x L_f = m_ice (334000)

3) Temperature of water rising from 0^oC to 29.3^oC = Q = m_icex c_water x (T_f-T_i) = m_icex(4190)x(29.3-0) =

= 122,767 J/Kg x m_ice

Therefore, total heat released by ice should be equal to that absorbed by water,

Q_{ice =} Q_water

=> m_ice x c_ice x (T_f-T_i) + m_ice x L_f + m_icex c_water x (T_f-T_i) = m_wter x c_wter x (Tf - Ti)

=> 30,870 J/kg x m_ice + m_ice (334000) + 122,767 J/Kg x m_ice = (0.350) x (4190) x(69.4-29.3)

=> m_ice = (58,806.65)/ (487637) = 0.1206 kg