Question

Four ice cubes at exactly 0 ∘C with a total mass of 53.5 g are combined with 130 g of water at 75 ∘C in an insulated container. (ΔH∘fus=6.02 kJ/mol, cwater=4.18J/g⋅∘C)

If no heat is lost to the surroundings, what is the final temperature of the mixture?

Express your answer using two significant figures.

Answer 1

At first, the four ice cubes at 0⁰ C will melt and will produce water at 0⁰ C.

The quantity of heat needed for the above mentioned process is Q = m₁ x ΔH_f (m₁ = mass of ice cubes)

After melting, the water at 0⁰ C warms in presence of hot water and at the same time the hot water cools down during the overall process (first melting of ice then to warm the water at 0⁰ C). The quantity of heat needed for this = m1 x ΔT x C_p .

Therefore the heat released by the hot water due to the overall process will be = m₂ x ΔT x C_p (m₂ = mass of hot water, ΔT is the change in temperature).

Now, let the final temperature be T_f.

Given, ΔH_f = 6.02 kJ/mol

= 6.02 x 10³J/mol = 6020 J/mol

So, for 18 g (1 mol) of water ΔH_f is 6020 J/mol

Therefore, for 1 g of water ΔH_f = 6020/18 = 334.44 J/g.

Therefore, total heat gained = total heat supplied (as not heat is lost)

or, (53.5 x 334.44) + [ (53.5 x (T_f - 0⁰) x 4.18] = [130 x (75⁰ - T_f) x 4.18]

or, 17892.54 + 223.63T_f = 543.4 x (75⁰ - T_f)

or, 17892.54 + 223.63T_f = 40755 - 543.4T_f

or, 223.63T_f + 543.4T_f = 40755 - 17892.54

or, 767.03T_f = 22862.46

or, T_f = 22862.46 / 767.03

or, T_f = 29.80

Answer: The final temperature of the mixture will be 29.80⁰ C.

[Hope this helps :)]