Question

An insulated beaker with negligible mass contains liquid water with a mass of 0.285 kg and a temperature of 81.6 ∘C .

How much ice at a temperature of -23.6 ∘C must be dropped into the water so that the final temperature of the system will be 40.0 ∘C ?

Take the specific heat of liquid water to be 4190 J/kg⋅K , the specific heat of ice to be 2100 J/kg⋅K , and the heat of fusion for water to be 3.34×10⁵ J/kg .

Answer 1

Heat lost by water = Heat gain by ice

m_water C_waterT = m_ice C_iceT + m_iceH + m_ice C_waterT_system

m_water C_water (T_i - T_f) = m_ice [C_ice (T_p - T') + H + C_water (T_sys,f - 0)]

where, m_water = mass of water = 0.285 kg

C_water = specific heat of liquid water = 4190 J/kg ⁰C

T_i = initial temperature of water = 81.6 ⁰C

T_f = final temperature of water = 0 ⁰C

T' = initial temperature of ice = - 23.6 ⁰C

C_ice = specific heat of ice = 2100 J/kg ⁰C

T_p = Freezing point of water = 0 ⁰C

H = latent heat of fusion for water = 3.34 x 10⁵ J/kg

T'_f = final temperature of the system = 40 ⁰C

then, we get

(0.285 kg) (4190 J/kg ⁰C) [(81.6 ⁰C) - (0 ⁰C)] = m_ice {(2100 J/kg ⁰C) [(0 ⁰C) + (23.6 ⁰C)] + (3.34 x 10⁵ J/kg) + (4190 J/kg ⁰C) [(40 ⁰C) - (0 ⁰C)]

(97442.6 J) = m_ice [(49560 J/kg) + (334000 J/kg) + (167600 J/kg)]

(97442.6 J) = m_ice (551160 J/kg)

m_ice = [(97442.6 J) / (551160 J/kg)]

m_ice = 0.176 kg