Question

How much ice at a temperature of -23.8 ∘C∘C must be dropped into the water so that the final temperature of the system will be 35.0 ∘C∘C ? Take the specific heat of liquid water to be 4190 J/kg⋅KJ/kg⋅K , the specific heat of ice to be 2100 J/kg⋅KJ/kg⋅K , and the heat of fusion for water to be 3.34×105 J/kgJ/kg .

Answer 1

0.0777 kg

Step-by-step explanation:

Answer 2

m_ice = 0.0777 kg

Step-by-step explanation:

Given

m_w = 0.22 kg

T_w,i = 79.7 C

T_ice,i = -23.8 C

T_eq = 35 C

C_w = 4190 J/kgK

C_ice = 2100 J/kgK

u_f = 334000 J/kg

The corresponding changes in beaker are:

water: Cools down dT = T_w,i - T_eq = 79.7 - 35 = 44.7 C

Q_loss = m_w * C_w * dT = 0.22*4190*44.7 = 41204.46 J

ice: Heats dT_1 = 0 + 23.8 = 23.8 C , phase change , Heats dT_2 = 35 - 0 = 35 C

Q_absorbed = (m_ice * C_ice * dT_1) + (m_ice*u_f) + (m_ice * C_w * dT_2)

Q_absorbed = m_ice * (2100*23.8 +334000 +4190*35) = 530630 *m_ice

Using Energy balance:

Q_loss = Q_absorbed

41204.46 = 530630 * m_ice

m_ice = 0.0777 kg