Answer:
0.102 cal/g.°C
Explanation:
According to the law of conservation of energy, the sum of the heat released by the electric burner (Qb) and the heat absorbed by the water (Qw) is zero.
Qb + Qw = 0
Qb = -Qw
Both heats can be calculated using the following expression.
Q = c × m × ΔT
where,
c: specific heat
m: mass
ΔT: change in the temperature
Then,
Qb = -Qw
cb × mb × ΔTb = - cw × mw × ΔTw
cb = - cw × mw × ΔTw / mb × ΔTb
cb = - (1 cal/g.°C) × 552.0 g × (80.3°C - 23.2°C) / 683.0 g × (23.2°C - 477.6°C)
cb = 0.102 cal/g.°C
