Water flow rate is $3.0\,\text{L min}^{-1}=3.0\,\text{kg min}^{-1}=0.05\,\text{kg s}^{-1}$. Heat needed per second is $\dot Q=\dot m c\Delta T=0.05(4200)(50)=1.05\times10^4\,\text{J s}^{-1}$. Fuel consumption rate $=\dot Q/(4.0\times10^4)=0.2625\,\text{g s}^{-1}$.
$0.26\,\text{g s}^{-1}$, or about $16\,\text{g min}^{-1}$.
Nitrogen is diatomic, so at room temperature $C_p=\frac72R=29.05\,\text{J mol}^{-1}\text{K}^{-1}$. Number of moles is $n=20\,\text{g}/28\,\text{g mol}^{-1}=0.714$. Therefore $Q=nC_p\Delta T=0.714(29.05)(45)=9.34\times10^2\,\text{J}$.
$9.3\times10^2\,\text{J}$ approximately.
Thermal equilibrium is set by heat lost by the hotter body equalling heat gained by the colder body; unequal masses or specific heats shift the final temperature away from the arithmetic mean. A coolant with high specific heat removes large heat without becoming too hot. During driving, flexing and friction heat the tyre and air, and at nearly fixed volume pressure rises with temperature. Coastal water heats and cools slowly, moderating harbour climates.
(a) Final temperature depends on heat capacities, not just the two temperatures. (b) High specific heat lets the coolant absorb much heat for a small temperature rise. (c) Tyre air warms during driving, increasing pressure. (d) Water has high specific heat, so the sea moderates temperature changes near harbours.
The compression is adiabatic because the system is insulated. For hydrogen, a diatomic gas, $\gamma=C_p/C_v=7/5=1.4$. Since $PV^\gamma$ is constant, $P_2/P_1=(V_1/V_2)^\gamma=2^{1.4}=2.64$.
By a factor $2^{1.4}\approx2.64$.
For the adiabatic process, $Q=0$ and work done on the system is $22.3\,\text{J}$, so work done by the system is $-22.3\,\text{J}$. From $Q=\Delta U+W$, $\Delta U=22.3\,\text{J}$. Internal energy change is state-dependent only, so it is the same for the second path. Heat absorbed is $9.35\times4.19=39.18\,\text{J}$. Hence $W=Q-\Delta U=39.18-22.3=16.9\,\text{J}$.
$16.9\,\text{J}$.
This is free expansion into vacuum in an insulated system. No heat is exchanged and no external work is done, so $\Delta U=0$. For an ideal gas, internal energy depends only on temperature, so $\Delta T=0$. The final volume is twice the initial volume at the same temperature, so pressure becomes half. During the sudden expansion the gas is not in thermodynamic equilibrium, so intermediate states are not represented on the equilibrium $P$-$V$-$T$ surface.
(a) Final pressure is half the initial pressure, i.e. $0.5\,\text{atm}$, in both cylinders. (b) Zero. (c) Zero for an ideal gas. (d) No.
By the first law, $\dot Q=dU/dt+\dot W$. Therefore $dU/dt=100-75=25\,\text{J s}^{-1}$.
$25\,\text{J s}^{-1}$.
From the graph, $D=(2.0\,\text{m}^3,600\,\text{N m}^{-2})$, $E=(5.0\,\text{m}^3,300\,\text{N m}^{-2})$, and $F=(2.0\,\text{m}^3,300\,\text{N m}^{-2})$. Work from D to E is the trapezium area: $W_{DE}=\frac{600+300}{2}(5-2)=1350\,\text{J}$. Work from E to F at constant pressure is $W_{EF}=300(2-5)=-900\,\text{J}$. Total work $=1350-900=450\,\text{J}$.
$450\,\text{J}$.