반응형 일반화학/[05장] 열화학384 125 g water 283 g 3.0℃ 91.0℃ 125 g water 283 g 3.0℃ 91.0℃ A 125 g sample of cold water and a 283 g sample of hot water are mixed in an insulated thermos bottle and allowed to equilibrate. If the initial temperature of the cold water is 3.0℃, and the initial temperature of the hot water is 91.0℃, what will be the final temperature? --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/.. 2023. 6. 16. 구리의 몰열용량. 226 kJ heat Cu 47.0 kg 12.5℃ 구리의 몰열용량. 226 kJ heat Cu 47.0 kg 12.5℃ If 226 kJ of heat increases the temperature of 47.0 kg of copper by 12.5℃, what is the molar heat capacity of copper? --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt 로부터, C = q / [m Δt] = (226 kJ) / [47.0 kg × 12.5 ℃] = 0.385 kJ/kg•℃ = 0.385 J.. 2023. 6. 16. 생성엔탈피로 반응엔탈피 계산. CH4(g) 불완전 연소열 생성엔탈피로 반응엔탈피 계산. CH4(g) 불완전 연소열 ΔH 계산. 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) --------------------------------------------------- CH4 불완전 연소 반응식 2CH4(g) + 3O2(g) → 2CO(g) + 4H2O(l) ( 참고 https://ywpop.tistory.com/11595 ) ΔH°_rxn = [생성물들의 ΔH°_f 합] – [반응물들의 ΔH°_f 합] ( 참고 https://ywpop.tistory.com/3431 ) = [2(–110.525) + 4(–285.8)] – [2(–74.9) + 3(0)] ( 참고: ΔH°_f 값 https://en.wikipedia.org/wiki/Standar.. 2023. 6. 16. ΔE 계산. 1.90 atm 8.30 L 2.80 L 350.0 J heat internal energy ΔE 계산. 1.90 atm 8.30 L 2.80 L 350.0 J heat internal energy A piston is compressed from a volume of 8.30 L to 2.80 L against a constant pressure of 1.90 atm. In the process, there is a heat gain by the system of 350.0 J. Calculate the internal energy change. --------------------------------------------------- w = –PΔV ( 참고 https://ywpop.tistory.com/5281 ) = –(1.90) (2.80 – 8.30) = +10.45 atm•L (+.. 2023. 6. 14. gold 55.5 g temp 20℃ 45℃ increase heat gold 55.5 g temp 20℃ 45℃ increase heat What quantity of heat is required to raise the temperature of 55.5 g of gold from 20℃ to 45℃? The specific heat of gold is 0.13 J/g•℃. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt = (0.13 J/g•℃) (55.5 g) (45 – 20 ℃) = (0.13) (55.5) (45 – 20.. 2023. 6. 11. Hg 25 g 100. J heat energy temp increase Hg 25 g 100. J heat energy temp increase If 100 J of heat energy is applied to a 25 g sample of mercury, by how many degrees will the temperature of the sample of mercury increase? The specific heat of mercury is 0.14 J/g•℃. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt 이므로, Δt =.. 2023. 6. 11. Al 42.7 g 15.2℃ heat energy Al 42.7 g 15.2℃ heat energy What quantity of heat energy must have been applied to a block of aluminum weighing 42.7 g, if the temperature of the block of aluminum increased by 15.2℃? The specific heat of aluminum is 0.89 J/g•℃. --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- q = C m Δt = (0... 2023. 6. 11. –30.℃ 얼음 1.00 mol을 140.℃ 수증기로 (kJ) –30.℃ 얼음 1.00 mol을 140.℃ 수증기로 (kJ) --------------------------------------------------- ▶ 참고: 열용량과 비열 [ https://ywpop.tistory.com/2897 ] --------------------------------------------------- H2O의 몰질량 = 18.015 g/mol 이므로, ▶ 얼음의 비열 = (2.03 J/℃•g) (18.015 g/mol) = 36.6 J/℃•mol ▶ 물의 비열 = (4.18 J/℃•g) (18.015 g/mol) = 75.3 J/℃•mol ▶ 수증기의 비열 = (2.02 J/℃•g) (18.015 g/mol) = 36.4 J/℃•mol ▶ 용융열 = 6.02 kJ/mol.. 2023. 6. 11. 생성엔탈피로 반응엔탈피 계산. CuO 연소열 생성엔탈피로 반응엔탈피 계산. CuO 연소열 CuO(s) + H2(g) → Cu(s) + H2O(l) --------------------------------------------------- ΔH°_rxn = [생성물들의 ΔH°_f 합] – [반응물들의 ΔH°_f 합] ( 참고 https://ywpop.tistory.com/3431 ) = [(0) + (–285.8)] – [(–155.2) + (0)] = –130.6 kJ ---> CuO(s) 1 mol당 반응열이므로, = –130.6 kJ/mol 답: ΔH° = –130.6 kJ/mol 2023. 6. 11. 중화열 계산. 0.500 M HCl + 0.500 M NaOH 1.00×10^2 mL 중화열 계산. 0.500 M HCl + 0.500 M NaOH 1.00×10^2 mL 일정 압력 열량계에서 0.500 M HCl 1.00×10^2 mL와 0.500 M NaOH 1.00×10^2 mL를 혼합하였다. HCl 용액과 NaOH 용액의 초기 온도는 22.50℃이었고, 혼합 용액의 최종 온도는 25.86℃이었다. 몰당 중화열을 계산하라. 용액의 밀도와 비열은 물의 값과 같다고 가정한다. (각각 1.00 g/mL, 4.184 J/g•℃) 열량계의 열용량은 무시한다. --------------------------------------------------- 1.00×10^2 mL = 100. mL = 100. g 1) 용액이 흡수한 열 q = C m Δt ( 참고 https://ywpop.tisto.. 2023. 6. 10. 이전 1 2 3 4 5 6 ··· 39 다음 반응형