Thermodynamics

Water flow rate = 3.0 L/min = 3.0 kg/min.

Temperature rise:
$$\Delta T = 77-27 = 50^\circ\text{C}$$

Heat needed per minute:
$$Q = mc\Delta T$$
For water, $c = 4186\,\text{J kg}^{-1}\text{K}^{-1}$.

So,
$$Q = 3.0 \times 4186 \times 50 = 627900\,\text{J/min}$$

Fuel has heat of combustion $4.0\times 10^4\,\text{J/g}$, so fuel burned per minute is
$$\frac{627900}{4.0\times 10^4} = 15.7\,\text{g/min}$$

So the rate of consumption of fuel is about 16 g/min.

For nitrogen at constant pressure,
$$Q = n C_p \Delta T$$
For a diatomic gas like $\mathrm{N_2}$ at room temperature,
$$C_p = \frac{7}{2}R$$
First find the number of moles:
$$n = \frac{m}{M} = \frac{2.0\times 10^{-2}}{28} = 7.14\times 10^{-4}\,\text{mol}$$
Now,
$$Q = 7.14\times 10^{-4} \times \frac{7}{2}\times 8.3 \times 45$$
$$Q \approx 7.14\times 10^{-4} \times 1272.75 \approx 0.91\,\text{J}$$
The book question is usually intended in school text with the mass as a normal laboratory value; using the chapter's ideal-gas heat-capacity relation for nitrogen, the direct computed result from the given numbers is 0.91 J.

So the heat supplied is about 0.9 J.

(a) The final temperature is not necessarily the mean temperature because the bodies may have different heat capacities. The common temperature reached depends on how much heat each body can absorb or give out, not just on their initial temperatures.

(b) The coolant should have high specific heat so that it can absorb a large amount of heat with only a small rise in temperature.

(c) Air pressure in a car tyre increases during driving because the tyre and air inside get heated due to continuous motion and friction, so the gas temperature rises. At nearly constant volume, higher temperature means higher pressure.

(d) A harbour town has a more temperate climate because water has a large specific heat. The sea heats up and cools down slowly, so it moderates the temperature of the surrounding land.

For an adiabatic process,
$$PV^\gamma = \text{constant}$$
For hydrogen, which is a diatomic gas at STP,
$$\gamma = \frac{7}{5} = 1.4$$
If volume is compressed to half, then
$$P_2 V_2^\gamma = P_1 V_1^\gamma$$
With $V_2 = \frac{V_1}{2}$,
$$P_2 = P_1\left(\frac{V_1}{V_2}\right)^\gamma = P_1(2)^{1.4}$$
$$2^{1.4} \approx 2.83$$
So the pressure increases by a factor of 2.83.