Numa máquina térmica uma parte da energia térmica fornecida ao sistema(Q1) é transformada em trabalho mecânico (τ) e o restante (Q2) é dissipado, perdido para o ambiente.
sendo:
τ: trabalho realizado (J) [Joule]
Q1: energia fornecida (J)
Q2: energia dissipada (J)
temos: τ = Q1 - Q2
O rendimento (η) é a razão do trabalho realizado pela energia fornecida:
η= τ/Q1
Exercícior resolvido:
Uma máquina térmica cíclica recebe 5000 J de calor de uma fonte quente e realiza trabalho de 3500 J. Calcule o rendimento dessa máquina térmica.
solução:
τ=3500 J
Q1=5000J
η= τ/Q1
η= 3500/5000
η= 0,7 ou seja 70%
Energia dissipada será:
τ = Q1 - Q2
Q2 = Q1- τ
Q2=5000-3500
Q2= 1500 J
Exercicio: Qual seria o rendimento se a máquina do exercicio anterior realizasse 4000J de trabalho com a mesma quantidade de calor fornecida ? Quanta energia seria dissipada agora?
obs: Entregar foto da resolução ou o cálculo passo a passo na mensagem
Answer:
0.00125 moles H₃X
Solution and Explanation:
In this question we are required to calculate the number of moles of triprotic acid neutralized in the titration.
Volume of NaOH used = final burette reading - initial burette reading
= 39.18 ml - 3.19 ml
= 35.99 ml or 0.03599 L
Step 1: Moles of NaOH used
Number of moles = Molarity × Volume
Molarity of NaOH = 0.1041 M
Moles of NaOH = 0.1041 M × 0.03599 L
= 0.00375 mole
Step 2: Balanced equation for the reaction between triprotic acid and NaOH
The balanced equation is;
H₃X(aq) + 3NaOH(aq) → Na₃X(aq) + 3H₂O(l)
Step 3: Moles of the triprotic acid (H₃X used
From the balanced equation;
1 mole of the triprotic acid reacts with 3 moles of NaOH
Therefore; the mole ratio of H₃X to NaOH is 1 : 3.
Therefore;
Moles of Triprotic acid = 0.00375 mole ÷ 3
= 0.00125 moles
Hence, moles of triprotic acid neutralized during the titration is 0.00125 moles.
Answer:
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Answer:
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Explanation: