Question

Объемгазапринагреванииувеличилсяна0,02м^3, авнутренняяэнергиявозрослана1280Дж. Какоеколичествотеплотыбылопереданогазу, еслипроцесспротекалприпостоянномдавлении на 1,5*10^5Па?

Answer 1

U= Q-A
Q=U+A;    A=pV
Q=U+pV
Q=1280+ 150000*0.02=4280(Дж)

Answer 2

Calculating the Heat Transferred

The question provides the following information: - The volume of gas increased by 0.02 m^3 - The internal energy of the gas increased by 1280 J - The process occurred at a constant pressure of 1.5*10^5 Pa

To calculate the heat transferred, we can use the first law of thermodynamics: Q = ΔU + W Where: Q is the heat transferred ΔU is the change in internal energy W is the work done

Since the process was at constant pressure: W = -PΔV Where: P is the pressure (1.5*10^5 Pa) ΔV is the change in volume (0.02 m^3)

Substituting the values given: ΔU = 1280 J ΔV = 0.02 m^3 P = 1.5*10^5 Pa

Calculating the work: W = -PΔV = -1.5*10^5 Pa * 0.02 m^3 = -3000 J

Using the first law: Q = ΔU + W = 1280 J + (-3000 J) = 4280 J

Therefore, the heat transferred to the gas was 4280 J.