1.0 L water Al pan 0.20 kg 20℃ boil energy heat 0.90 J/g-K

1.0 L water Al pan 0.20 kg 20℃ boil energy heat 0.90 J/g-K

1.0 L 물이 담긴 알루미늄 프라이팬(0.20 kg)을 가열한다면, 상온(20 ℃)에서 끓을 때까지 얼마나 많은 에너지(열)가 필요한가? 팬과 물 사이에 열적 평형은 항상 유지된다고 가정한다(Al과 H2O의 비열은 0.90 J/g-K과 4.18 J/g-K이다. 물의 밀도는 1 g/mL이다).

How much energy (heat) does it take to heat an Al saucepan (0.20 kg) containing 1.0 L of water from room temperature (20 ℃) to boiling? Assume thermal equilibrium between the pan and water at all times (specific heat capacity for Al is 0.90 J/g-K and H2O is 4.18 J/g-K. Assume the density of water is 1 g/mL).

물 가열에 필요한 열을 계산하면,

q = C m Δt

( 참고 https://ywpop.tistory.com/2897 )

= (4.18 J/g•℃) (1000 g) (100 – 20 ℃)

= 334400 J

= 334.4 kJ

알루미늄 가열에 필요한 열을 계산하면,

q = C m Δt

= (0.90 J/g•℃) (200 g) (100 – 20 ℃)

= 14400 J

= 14.4 kJ

334.4 + 14.4 = 348.8 kJ

답: 약 350 kJ

[키워드] If you heat an aluminum frying pan (0.20 kg) containing 1.0 L of water, how much energy (heat) is needed to bring it to a boil at room temperature (20 ℃)? It is assumed that thermal equilibrium between the pan and water is always maintained (the specific heats of Al and H2O are 0.90 J/g-K and 4.18 J/g-K; the density of water is 1 g/mL).