If a refrigerator's door is kept open, will the room become cool or hot? Explain.
If a refrigerator's door is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Yes, during adiabatic compression the temperature of a gas increases while no heat is given to it.
$$\begin{aligned} \text{In adiabatic compression,}\quad & d Q=0 \\ \therefore \quad \text{From first law of thermodynamics,}\quad & d U=d Q-d W \\ & d U=-d W \end{aligned}$$
In compression work is done on the gas i.e., work done is negative.
Therefore, $$\quad d U=\text { Positive }$$
Hence, internal energy of the gas increases due to which its temperature increases.
Air pressure in a car tyre increases during driving. Explain.
During, driving, temperature of the gas increases while its volume remains constant.
So, according to Charle's law, at constant volume ($V$),
Pressure $(p) \propto$ Temperature $(T)$
Therefore, pressure of gas increases.
Consider a Carnot's cycle operating between $T_1=500 \mathrm{~K}$ and $T_2=300 \mathrm{~K}$ producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
$$\begin{aligned} \text{Given,}\quad \text { temperature of the source } T_1 & =500 \mathrm{~K} \\ \text { Temperature of the sink } T_2 & =300 \mathrm{~K} \\ \text { Work done per cycle } W & =1 \mathrm{~kJ}=1000 \mathrm{~J} \end{aligned}$$
Heat transferred to the engine per cycle $Q_1=$ ?
Efficiency of a Carnot engine $(\eta)=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}$
$$\text{and}\quad \eta=\frac{W}{Q_1}$$
$$\Rightarrow \quad Q_1=\frac{W}{\eta}=\frac{1000}{(2 / 5)}=2500 \mathrm{~J}$$
A person of mass 60 kg wants to lose 5 kg by going up and down a 10 m high stairs. Assume he burns twice as much fat while going up than coming down. If 1 kg of fat is burnt on expending 7000 kcal , how many times must he go up and down to reduce his weight by 5 kg ?
Given, $\quad$ height of the stairs $=h=10 \mathrm{~m}$
Energy produced by burning 1 kg of fat $=7000 \mathrm{~kcal}$
$\therefore$ Energy produced by burning 5 kg of fat $=5 \times 7000=35000 \mathrm{~kcal}$ $=35 \times 10^6 \mathrm{~cal}$
$$\begin{aligned} &\text { Energy utilised in going up and down one time }\\ &\begin{aligned} & =m g h+\frac{1}{2} m g h=\frac{3}{2} m g h \\ & =\frac{3}{2} \times 60 \times 10 \times 10 \\ & =9000 \mathrm{~J}=\frac{9000}{4.2}=\frac{3000}{1.4} \mathrm{cal} \end{aligned} \end{aligned}$$
$$\begin{aligned} &\therefore \text { Number of times, the person has to go up and down the stairs }\\ &\begin{aligned} & =\frac{35 \times 10^6}{(3000 / 1.4)}=\frac{35 \times 1.4 \times 10^6}{3000} \\ & =16.3 \times 10^3 \text { times } \end{aligned} \end{aligned}$$