According to the first law of thermodynamics, the change in internal energy is:

$$
Delta E=Q-W
$$

where $Q$ is the heat and $W$ is the work.

The aim of pushing the plunger rapidly is to create an adiabatic process, in an adiabatic process, there is no flow of energy in or out of the system thanks to good insolation, or in our case, the process happens so fast that the heat does not have time to flow.

This simply means $Q=0 Rightarrow Delta E=-W$

Since we are pushing the plunger then we are doing work on the system and $Delta E>0$

The dramatic increase in thermal energy creates another dramatic increase in temperature which ignites the bits of paper.

$$
tt{(a) According to the first law of thermodynamics, the change in internal energy is \
$$Delta E=Q-W$$}
$$

where $Q$ is the heat and $W$ is the work.

For an isothermal process, there is no change in internal energy:

$$
Delta E=Q-W=0Rightarrow Q=Wtext{ (1)}
$$

In order for the gas to expand, it needs to do work, we can deduce from (1) that this amount of work is drawn in the form of heat from the surroundings $Q>0$.

Finally based on the entropy definition:

$$
Delta s=frac{Q}{T}
$$

We can see that $Q$ and $Delta s$ are proportional and since $Q>0$ and $T>0$ the change of entropy is positive $Delta s>0$

The entropy of the universe increases.

(b) Explanation B

$$
tt{(a) The entropy increase, (b) B}
$$

$tt{Efficiency of a Heat Engine equation:}$

$$
e=frac{W}{Q_h}
$$

with $W$ being the work done by the heat engine, and $Q_h$ the heat supplied to engine.

if we suppose that the $Q_h=4W$, then we can rewrite the efficiency equation as:

$$
e=frac{W}{4W}=frac{1}{4}
$$

$$
e=frac{1}{4}
$$

$tt{(a) The efficiency of a Heat Engine equation:}$

$$
e=frac{W}{Q_h}
$$

with $W$ being the work done by the heat engine, and $Q_h$ the heat supplied to the engine.

if we suppose that $Q_h=5W$, then we can rewrite the efficiency equation as:

$$
e=frac{W}{5W}=frac{1}{5}
$$

According to Carnot’s Theorem of Maximum Efficiency:

$$
e=1-frac{T_C}{T_h}
$$

with $T_C$ being the temperature of the cold reservoir and $T_h$ the temperature of the hot reservoir.

having $T_h=490k$ we can deduce $T_c$:

$$
T_c=T_h(1-e)=490(1-frac{1}{5})=boxed{392k}
$$

(b) we know that:

$$
e=frac{W}{Q_h}
$$

$$
e=frac{Q_h-Q_C}{Q_h}
$$

$$
e=1-frac{Q_C}{Q_h}
$$

$$
Rightarrow Q_c=(1-e)Q_h=boxed{0.8Q_h}
$$

$$
tt{(a) $T_c=392k$, (b) $Q_c=0.8Q_h$}
$$

$tt{(a) As heat is being added to the Ice$Q>0$, fusion occurs and the molecules are now in disorder and free to move(in liquid form) rather than being confined to a specific space, therefore entropy increases }$

(b) The Definition of Entropy Change is :

$$
Delta s =frac{Q}{T}
$$

with $Q$ being the heat added/removed from the system and $T$ the temperature in kelvin.

Since water is in a phase transition, we need to calculate the heat needed to transform it fully into a liquid:

$$
Q=L_fm
$$

with $L_f$ being the fusion latent heat of the water and $m=0.13kg$ the mass of ice melted:

$$
Q_T=33.5*10^4*0.13=4.355*10^4J
$$

We can now calculate the exact change in entropy:

$$
Delta s= frac{4.355*10^4}{273.15}=boxed{159.44frac{J}{k}}
$$

$$
tt{(a) entropy increase, (b) $Delta s =159.44frac{J}{k}$ }
$$

$tt{(a) Efficiency of a Heat Engine equation:}$

$$
e=frac{W}{Q_h}
$$

with $W$ being the work done by the heat engine, and $Q_h$ the heat supplied to the engine.

We know that the engine gets a supply of $Q_h=1250J$ and performs work $W=1120J$, this yields:

$$
e=frac{1120}{1250}=boxed{0.896}
$$

According to Carnot’s Theorem of Maximum Efficiency:

$$
e=1-frac{T_C}{T_h}=1-frac{302}{1010}=boxed{0.7}
$$

The efficiency claimed by the inventor is way higher than the one calculated based on values of the temperature, one should not invest in such an invention.

(a) $e=0.896$, (b) $e=0.7$, one should not invest

(a) as the water freezes at night, it is losing heat to the surroundings, therefore $Q<0$, on the other hand, the change of entropy is defined as:

$$
Delta s =frac{Q}{T}
$$

With $Q$ being the heat supplied/removed and $T$ the temperature in kelvin.

We can deduce from the equation that since $Q0$ that the entropy of the water decreases $Delta s <0$.

(b) First we have to remember the formula corresponding to the heat required to change the water from water to ice :

$$
Q=mL_f=0.53*33.5*10^4=17.755*10^4text{ J}
$$

Then we can calculate the change in entropy by simply applying the formula:

$$
Delta s =frac{-17.755*10^4}{273.15-0.5}=boxed{-651.2frac{text{J}}{text{k}}}
$$

(c) As the dish is losing heat, its surroundings are gaining that heat, and its entropy is increasing.

$$
tt{(a) decrease, (b) $Delta s =-652.2frac{text{K}}{text{j}}$, (c) The surrondings entropy increases}
$$

Carnot’s Theorem of Maximum Efficiency $e$ states:

$$
e=1-frac{T_C}{T_h}
$$

where $T_C$ is the temperature of the cold reservoir and $T_h$ is the temperature of the hot reservoir
We will consider any change including the example of $50k$ as $Delta T$

(a) raising the temperature of $T_h$ to $T_h+Delta T$ means:

$$
e_1=1-frac{T_C}{T_h+Delta T}=frac{T_h+Delta T-T_C}{T_h+Delta T}text{ (1)}
$$

(b) Lowering the temperature of $T_C$ to $T_C-Delta T$ means:

$$
e_2=1-frac{T_C-Delta T}{T_h}=frac{T_h-T_C+Delta T}{T_h}text{ (2)}
$$

Deviding (1) per (2) yields:

$$
frac{e_1}{e_2}=frac{T_h}{T_h+Delta T}<1
$$

$$
Rightarrow e_1<e_2
$$