Specific Heat Question

Question

Given 350.5 g of hot tea at 70.0 ∘C, what mass of ice at 0 ∘C must be added to obtain iced tea at 13.0 ∘C? The specific heat of the tea is 4.18 J/(g⋅∘C), and ΔHfusion for ice is +6.01 kJ/mol

Answer 1

To get the 350.5 g of hot tea @ 70º to cool down to 13.0º, obviously it has to lose heat. That heat that is lost will go to the cold ice @ 0º and the temperature of the ice will rise to 13º (thermal equilibrium). According to the law of conservation of energy, the heat LOST by the hot tea must be equal to the heat GAINED by the cold ice.

We will use q = mC∆T to find the heat lost by the hot tea.

q = heat lost = ?

m = mass = 350.5 g

C = specific heat = 4.18 J/gº

∆T = change in temperature = 70 - 13 = 57º

q = (350.5 g)(4.18 J/gº)(57º) = 83,510 J .. this is the heat lost by the ice (should actually have a - sign).

Next, we will find the mass of ice that would be required to gain this much heat. First, we will have to melt the ice @ 0º (phase change and no change in temperature) and then have the final temperature reach 13.0ºC

q = m∆H_fusion to melt the ice

q = mC∆T to change the temperature from 0º to 13.0º

q = m∆H_fusion + mC∆T

Since ∆H_fusion is given to us a 6.01 kJ/mol, we will convert it to J/g to be consistent with our other calcs.

6.01 kJ/mol x 1000 J/kJ x 1 mol/18.0 g = 334 J/g =∆H_fusion

83,510 J = (m)(334 J/g) + (m)(4.18 J/gº)(13º)

83,510 = 334m + 54.3m

83510 = 388m

m = 215 g of ice needed