Question

A 30.5 g sample of an alloy at 91.9°C is placed into 49.4 g water at 25.0°C in an insulated coffee cup. The heat capacity of the coffee cup (without the water) is 9.2 J/K. If the final temperature of the system is 31.1°C, what is the specific heat capacity of the alloy? (c of water is 4.184 J/g×K) ____J/g°C?

Answer 1

0.710J/g ℃

Explanation

Heat absorbed by Coffee cup is q_c

  q_c = ∆T × C

  _{∆T = Tempereture difference}

_{C = Heat capacity of Coffee Cup, 9.2J/K}

Initial temperature = 25℃=298.15K

Final temperature = 31.1℃= 304.25K

∆T =( 304.25 - 298.15 K) = 6.1K

Therefore,

q_c = 6.1K × 9.2J/K = 56.12J

Heat absorbed by water is q_w

q_w = m × ∆T × C

m = mass of water , 49.4g

∆T = 6.1K

C = specific heat capacity of water , 4.184 J/g K

q_w = 49.4g × 6.1K × 4.184(J/g K)

= 1260.8J

Total heat absorbed, q_abs = 1260.8J + 56.12J =1316.92J

Heat released by the alloy is q_rel

  q_rel = m × ∆T × C_a

m = mass of alloy = 30.5g

∆T = 304.25K - 365.05K = -60.8

C_a = Specific heat capacity of alloy

q_rel = 30.5g × (-60.8K) × C_a

   _{= -1854.4(gK)× Ca}

Heat absorbed (q_abs )= -Heat released (q_rel)

1316.9J = -(-1854.4(gK) × C_a)

  C_a =1316.9J/(1854.4(g K)

= 0.710J/g K

= 0.710J/g ℃