Question

When 27.2 g of solid A, at its melting point of 34.0 degree C, was added to 62.7 g of liquid A at 51.2 degree C, the mixture cooled to 35.0 degree C, with no solid remaining. Given the specific heat of liquid A (s = 0.798 J/g degree C), find the heat of fusion per gram of solid A. Assume that no heat enters or leaves the mixture. That is, assume that the mixture is an isolated system and therefore that q_1 + q_2 + q_3 = 0.

Answer 1

heat of fusion of solid A = 29.0 J/g

Explanation

Heat lost by hot liquid A = (mass of hot liquid A) * (specific heat of liquid A) * (temperature change for hot liquid A)

Heat lost by hot liquid A = (62.7 g) * (0.798 J/g.^oC) * (51.2 ^oC - 35.0 ^oC)

Heat lost by hot liquid A = 810.56052 J

Heat gained by cold liquid A = [(heat of fusion) * (mass of cold solid A)] + [(mass of cold liquid A) * (specific heat of liquid A) * (temperature change for cold liquid A)]

Heat gained by cold liquid A = [(heat of fusion) * (27.2 g)] + [(27.2 g) * (0.798 J/g.^oC) * (35.0 ^oC - 34.0 ^oC)]

Heat gained by cold liquid A = [(heat of fusion) * (27.2 g)] + 21.7056 J

Since heat energy is conserved,

Heat gained by cold liquid A = Heat lost by hot liquid A

[(heat of fusion) * (27.2 g)] + 21.7056 J = 810.56052 J

(heat of fusion) * (27.2 g) = 810.56052 J - 21.7056 J

(heat of fusion) * (27.2 g) = 788.85492 J

heat of fusion of solid A = 788.85492 J / 27.2 g

heat of fusion of solid A = 29.0 J/g