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1 answer:
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Answer:
146 kJ
Explanation:
There are two heat flows in this question.
Heat lost on cooling + heat lost on solidifying = 0
q₁ + q₂ = 0
mCΔT + nΔHsol = 0
Data:
m = 575 g
C = 0.449 J·K⁻¹g⁻¹
T_i = 1825 K
T_f = 1811 K
ΔHsol = -13.8 kJ·mol⁻¹
Calculations:
(a) Heat lost on cooling
ΔT = T_f - T_i = 1811 K - 1825 K = -14 K
q₁ = mCΔT = 575 g × 0.449 J·K⁻¹g⁻¹ × (-14 K) = -361 J = -3.61 kJ
(b) Heat lost on solidifying
(c) Total heat lost
q = q₁ + q₂ = -3.61 kJ - 142.1 kJ = -146 kJ
The heat lost was 146 kJ.