Answer to Question #220834 in Chemistry for LembuBrapi

Question

Calculate the initial temperature if 4.2 kg piece of gallium metal absorbed 630122 J of heat at 224.1 C. [Given specific heat, s, of Ga = 0.368 J g-1 C -1]

Answer 1

Solution:

q = m × S × ΔT

q = m × S × (T_f - T_i),

where:

q = amount of heat energy gained or lost by substance (J)

m = mass of sample (g)

S = specific heat (J °C^-1 g^-1)

T_f = final temperature (°C)

T_i = initial temperature (°C)

 

Thus:

(630122 J) = (4200 g) × (0.368 J °C^-1 g^-1) × (224.1 - T_i)°C

(630122) = (1545.6) × (224.1 - T_i)

224.1 - T_i = (630122) / (1545.6) = 407.7

T_i = 224.1 - 407.7 = -183.6

T_i = -183.6°C

But, melting point of gallium is 29.76°C.

Therefore, either a mistake was made in the problem statement, or the melting of solid gallium must be taken into account.

Latent heat of fusion of gallium (L_f) is 80.1 J g^-1

q = L_f × m + m × S × (T_f - T_i)

(630122 J) = (80.1 J g^-1) × (4200 g) + (4200 g) × (0.368 J °C^-1 g^-1) × (224.1 - T_i)°C

(630122) = (336420) + (1545.6) × (224.1 - T_i)

(293702) = (1545.6) × (224.1 - T_i)

224.1 - T_i = (293702) / (1545.6) = 190.0

T_i = 224.1 - 190.0 = 34.1

T_i = 34.1°C - it is more than the melting point of gallium (29.76°C).

Therefore, most likely the problem statement contains a mistake.