Molar specific heat capacity is defined as heat energy required to increase the temperature of one mole of substance by 1 K or 1⁰C
C = \(\frac { 1 }{ u } \left( \frac { \triangle Q }{ \triangle T } \right) \)
Here C is known as molar specific heat capacity of a substance and f. L is number of moles in the substance.

Here, the process is adiabatic compression. The volume is given and temperature is to be found. we can use the equation (8.38 )
\({ T }_{ i }{ V }_{ i }^{ \Upsilon -1 }{ =T }_{ f }{ V }_{ f }^{ \Upsilon -1 }.\)
T_i = 300 K (273 + 27°C = 300 K)
\({ V }_{ i }=V\& { V }_{ f }=\frac { V }{ 4 } \)
\({ T }_{ f }={ T }_{ i }{ \left( \frac { { V }_{ i } }{ { V }_{ f } } \right) }^{ \Upsilon -1 }\) = 300 K × 4^1.4-1 = 300K\(\times\)1.741
T2 ≈ 522 K or 2490C
This temperature is higher than the boiling point of water. So it is very dangerous to touch the nozzle of blocked pump when you pump air.

Consider a container of gas with volume V, pressure P and temperature T. If we add sand particles one by one slowly on the top of the piston, the piston will move inward very slowly. This can be taken as almost a quasi-static process. It is shown in the figure

Sand particles added slowly- quasi-static process

\(\frac { \Delta L }{ L } ={ \alpha }_{ L }{ \Delta }_{ T }\)
\(\Delta L={ L\alpha }_{ L }{ \Delta }_{ T }\)
\(\Delta\)L = 10\(\times\)10^-6 \(\times\)300\(\times\)23 = 0.69 m=69 mm
Area Expansion
For a small change in temperatur ΔT the fractional change in area \(\left(ΔA\over A_0\right)\) substance is directly proportional to ΔT and it can be written as
\({ΔA\over A_0}=\alpha_AΔT\)
Therefore, \(\alpha_A={ΔA\over A_0ΔT}\)
Where, α_A = coefficient of area expansion.
ΔA = Change in area
A₀ = Original area
ΔT = Change in temperature
Volume Expansion
For a small change in temperature ΔT the fractional change in volume \(\left(ΔV\over V_0\right)\) of a substance is directly proportional to ΔT.
\({ΔV\over V_0}=\alpha_VΔT\)
Therefore, \(\alpha_V={ΔV\over V_0ΔT}\)
Where, α_V = coefficient of volume expansion.
ΔV = Change in volume
V₀ = Original volume
ΔT = Change in temperature
Unit of coeffi cient of linear, area and volumetric expansion of solids is ^oC^-1 or K^-1

a. When it rains, lake receives water from the cloud. Once the rain stops, the lake will have more water than before raining. Here ‘raining’ is a process which brings water from the cloud. Rain is not a quantity rather it is water in transit. So the statement ‘lake has more rain is wrong, instead the ‘lake has more water will be appropriate.
b. When heated, a cup of coffee receives heat from the stove. Once the coffee is taken from the stove, the cup of coffee has more internal energy than before. ‘Heat’ is the energy in transit and which flows from an object at higher temperature to an object at lower temperature. Heat is not a quantity. So the statement ‘A hot cup of coffee has more heat is wrong, instead ‘coffee is hot’ will be appropriate.

Thermal expansion is the tendency of matter to change in shape, area, and volume due to a change in temperature.

Specific heat capacity of a substance is defined as the amount of heat energy required to raise the temperature of 1 kg of a substance by 1 Kelvin or 1^oC. Its unit is JKg^-1 K^-1.

The efficiency of heat engine is given by
\(\eta =1-\frac { { Q }_{ L } }{ { Q }_{ H } } \)
\(\eta =1-\frac { 300 }{ 500 } =1-\frac { 3 }{ 5 } \)
\(\eta \) = 1 – 0.6 = 0.4
The heat engine has 40% efficiency, implying that this heat engine converts only 40% of the input heat into work.

\(\mathrm{COP}=\frac{T_{L}}{T_{H}-T_{L}}\)
\(\mathrm{T}_{\mathrm{L}}=-12+273 =261 \mathrm{~K} \)
\(5=\frac{261}{T_{H}-261} \)
\(\therefore 5\left(T_{H}-261\right) =261 \)
\(5 \mathrm{~T}_{\mathrm{H}}-1305 =261 \)
\(5 \mathrm{~T}_{\mathrm{H}} =261+1305 =1566 \)
\(\therefore T_{H} =\frac{1566}{5} \)
\(=313.2 \mathrm{~K} \)
\(\mathrm{~T}_{\mathrm{H}}=313.2-273 =40.2^{\circ} \mathrm{C} \)

\(\mathrm{T}_{2} =0^{\circ} \mathrm{C}+273=273 \mathrm{~K} \)
\(\mathrm{~T}_{1}=100^{\circ} \mathrm{C}=100+273=373 \mathrm{~K} \)
\(\eta =1-\frac{T_{2}}{T_{1}} \)
\(=1-\frac{273}{373} \)
\(=\frac{373-273}{373} \)
\(=\frac{100}{373}=0.26809 \times 100 \)
\(=26.809 \% \)

\(\lambda_{m} T =\mathrm{b} \)
\(\therefore T =\frac{2.898 \times 10^{-3}}{350 \times 10^{-9}}\)
\(\mathrm{~T} =0.00828 \times 10^{6} \)
\(=8280 \mathrm{~K} \)

During the thermal equilibrium with surroundings internal energy is increased and heat energy will be increased.
\(\Delta U>0 Q>0\)