Answer:
(a) ΔU = 7.2x10²
(b) W = -5.1x10²
(c) q = 5.2x10²
Explanation:
From the definition of power (p), we have:
[tex] p = \frac {\Delta W}{\Delta t} = \frac {\Delta U}{\Delta t} [/tex] (1)
where, p: is power (J/s = W (watt)) W: is work = ΔU (J) and t: is time (s)
(a) We can calculate the energy (ΔU) using equation (1):
[tex] \Delta U = p \cdot \Delta t = 100 \frac{J}{s} \cdot 2.0\cdot 10^{-2} h \cdot \frac{3600s}{1h} = 7.2 \cdot 10^{2} J [/tex]
(b) The work is related to pressure and volume by:
[tex] \Delta W = -p \Delta V [/tex]
where p: pressure and ΔV: change in volume = V final - V initial
[tex] \Delta W = - p \cdot (V_{fin} - V_{ini}) = - 1.0 atm (5.88L - 0.85L) = - 5.03 L \cdot atm \cdot \frac{101.33J}{1 L\cdot atm} = -5.1 \cdot 10^{2} J [/tex]
(c) By the definition of Energy, we can calculate q:
[tex] \Delta U = \Delta W + \Delta q [/tex]
where Δq: is the heat transfer
[tex] \Delta q = \Delta U - \Delta W = 7.2 J - (-5.1 \cdot 10^{2} J) = 5.2 \cdot 10^{2} J [/tex]
I hope it helps you!
