5400kJ of heat energy was lost when some amount of steam condensed to water for drinking purposes at 15º C. What is the quantity of water collected? [L\(_f \) = 2.26 × 10\(^6\) Jkg\(^{-1}\), c\(_w\) = 4200 Jkg\(^{-1}K^{-1}\)]
Answer: D. 2.06 kg
(Steam at 100º C: heat lost = m(L\(_v\) + c_w(100 - 15)) = 5400 × 10\(^3\) J;
m ≈ \(\frac{5.4e6}{(2.26e6 + 4200 \times 85)}\) ≈ 2.06 kg.)
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