Question

8) A person makes ice tea by adding ice to 1.8 kg of hot tea, initially at 80°C. How many kilograms of ice, initially at 0.00°C, are required to bring the tea to 0°C? The heat of fusion of ice is 334 k]/kg, and we can assume that tea has essentially the same thermal properties as water, so its specific heat is 4190 J/(kg.K).

Answer 1

Heat lost by hot object = heat gained by cold object

Final temperature of hot object = final temperature of cold object. Given as 10 C.

Heat lost by hot object = m_w*c_w*(Tw1 - T2)

Heat gained by cold object = m_i*c_w*(T2 - Ti1) + m_i*hf

Remember. The ice experiences a change of phase in addition to a change in temperature. During its change in temperature, it has the same specific heat capacity as water.

Equate:

m_w*c_w*(Tw1 - T2) = m_i*c_w*(T2 - Ti1) + m_i*hf

Group m_i terms:

m_w*c_w*(Tw1 - T2) = m_i*(c_w*(T2 - Ti1) + hf)

Solve for m_i:

m_i = m_w*c_w*(Tw1 - T2) / (c_w*(T2 - Ti1) + hf)

Data:

m_w = 1.8 kg

c_w = 1 kcal/kg-C

hf = 80 kcal/kg

Tw1 = 80C

Ti1 = 0C

m_i = 1.8*1*(80 - 10) / (1*(10 - 0) + 80)

Result:

m_i = 1.4 kg