Refrigerator Questions in English

(D) The coefficient of performance $(\beta)$ of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir $(Q_2)$ to the work done on the system $(W)$: $\beta = \frac{Q_2}{W}$.
Since $W = Q_1 - Q_2$, we have $\beta = \frac{Q_2}{Q_1 - Q_2}$.
Dividing numerator and denominator by $Q_1$, we get $\beta = \frac{Q_2/Q_1}{1 - Q_2/Q_1}$.
Using the efficiency of a Carnot engine $\eta = 1 - \frac{Q_2}{Q_1}$, we can write $\frac{Q_2}{Q_1} = 1 - \eta$.
Substituting this into the expression for $\beta$: $\beta = \frac{1 - \eta}{\eta} = \frac{1}{\eta} - 1$.
Since the efficiency $\eta$ of any heat engine is always less than $1$, the term $\frac{1}{\eta}$ is always greater than $1$.
Therefore, $\beta = \frac{1}{\eta} - 1$ can take any positive value greater than $0$.
Since $1$, $0.5$, and $9$ are all positive values, all of these can be the coefficient of performance of a refrigerator.

(A) Given: Temperature inside the refrigerator $T_2 = 273 \, K$,Temperature outside (surrounding) $T_1 = 300 \, K$.
For a reversible refrigerator,the coefficient of performance $\beta$ is given by $\beta = \frac{T_2}{T_1 - T_2}$.
Substituting the values: $\beta = \frac{273}{300 - 273} = \frac{273}{27} \approx 10.11$.
The coefficient of performance is also defined as $\beta = \frac{Q_2}{W}$,where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done.
Given $W = 1 \, J$,we have $Q_2 = \beta \times W = 10.11 \times 1 = 10.11 \, J$.
The heat delivered to the surrounding $Q_1$ is given by $Q_1 = Q_2 + W$.
$Q_1 = 10.11 \, J + 1 \, J = 11.11 \, J$.
Rounding to the nearest integer,the heat delivered is approximately $11 \, J$.

(C) The correct option is $C$.
$A$ refrigerator works as a heat pump that extracts heat from its interior and rejects it into the surrounding environment (the room).
When the door is open,the refrigerator continues to extract heat from the room air and releases it back into the room along with the heat generated by the electrical work done by the compressor.
Since the compressor performs work on the system,the total energy released into the room is the sum of the heat extracted and the work done by the motor.
Therefore,the net effect is an increase in the total thermal energy of the room,which ultimately warms the room slightly.

(B) The coefficient of performance $(COP)$,$\beta$,for an ideal refrigerator is given by the formula: $\beta = \frac{T_2}{T_1 - T_2}$,where $T_2$ is the temperature of the freezer and $T_1$ is the temperature of the surroundings (sink).
Given: $\beta = 5$ and $T_2 = -13^{\circ} C = (-13 + 273) K = 260 K$.
Substituting the values into the formula:
$5 = \frac{260}{T_1 - 260}$
$5(T_1 - 260) = 260$
$5T_1 - 1300 = 260$
$5T_1 = 1560$
$T_1 = 312 K$
Converting back to Celsius: $T_1 = (312 - 273)^{\circ} C = 39^{\circ} C$.

(C) The coefficient of performance $\alpha$ of a refrigerator is defined as the ratio of heat extracted from the cold reservoir $(Q_2)$ to the work done $(W)$ on the system: $\alpha = \frac{Q_2}{W}$.
Since $W = Q_1 - Q_2$,where $Q_1$ is the heat released to the hot reservoir,we have $\alpha = \frac{Q_2}{Q_1 - Q_2}$.
Cross-multiplying gives: $\alpha(Q_1 - Q_2) = Q_2$.
Expanding the equation: $\alpha Q_1 - \alpha Q_2 = Q_2$.
Rearranging to solve for $Q_2$: $\alpha Q_1 = Q_2 + \alpha Q_2 = Q_2(1 + \alpha)$.
Therefore,$Q_2 = \frac{\alpha Q_1}{1 + \alpha}$.

(C) The coefficient of performance $(COP)$ of a refrigerator is given by the formula: $COP = \frac{T_L}{T_H - T_L}$,where $T_L$ is the temperature of the freezer and $T_H$ is the room temperature in Kelvin.
Given: $COP = 5$,$T_L = -13^{\circ} C = (-13 + 273) K = 260 K$.
Substituting the values into the formula: $5 = \frac{260}{T_H - 260}$.
Multiplying both sides by $(T_H - 260)$: $5(T_H - 260) = 260$.
$T_H - 260 = \frac{260}{5} = 52$.
$T_H = 260 + 52 = 312 K$.
Converting back to Celsius: $T_H = (312 - 273)^{\circ} C = 39^{\circ} C$.

(D) The coefficient of performance $(\beta)$ of a refrigerator is given by the formula: $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.
Given: $\beta = 5$ and $Q_2 = 250 \ J$.
Substituting the values: $5 = \frac{250}{W} \implies W = \frac{250}{5} = 50 \ J$.
The heat released to the room $(Q_1)$ is the sum of the heat extracted and the work done: $Q_1 = Q_2 + W$.
$Q_1 = 250 \ J + 50 \ J = 300 \ J$.
Therefore, the heat released per cycle to the room is $300 \ J$.

(B) The efficiency of a Carnot heat engine is given by $\eta = 10 \% = 0.1$.
When the same engine operates as a refrigerator,the coefficient of performance $(COP)_R$ is related to the efficiency $\eta$ of the heat engine by the formula:
$(COP)_R = \frac{1 - \eta}{\eta}$.
Substituting the value of $\eta = 0.1$ into the formula:
$(COP)_R = \frac{1 - 0.1}{0.1} = \frac{0.9}{0.1} = 9$.
Thus,the coefficient of performance of the refrigerator is $9$.

(B) The rate of heat entering the cold storage is $Q_2 = \frac{mL}{t} = \frac{2 \times 10^3 \ g \times 80 \ cal/g}{3600 \ s} = \frac{160000}{3600} \ cal/s = \frac{400}{9} \ cal/s$.
Converting to Joules per second (Watts): $Q_2 = \frac{400}{9} \times 4.2 \ J/s = 186.67 \ W$.
The coefficient of performance $(COP)$ of a Carnot refrigerator is given by $\text{COP} = \frac{T_2}{T_1 - T_2} = \frac{Q_2}{W}$.
Here,$T_2 = 0^{\circ} C = 273 \ K$ and $T_1 = 20^{\circ} C = 293 \ K$.
So,$\frac{273}{293 - 273} = \frac{186.67}{W}$.
$\frac{273}{20} = \frac{186.67}{W}$.
$W = \frac{186.67 \times 20}{273} \approx 13.67 \ W$.
Thus,the minimum power output is approximately $13.6 \ W$.

(C) The coefficient of performance $(\beta)$ of a refrigerator is defined as $\beta = \frac{Q_2}{W}$, where $Q_2$ is the heat extracted from the cold reservoir and $W$ is the work done on the system.
Given $\beta = 0.25 = \frac{1}{4}$, so $Q_2 = \frac{W}{4}$.
The heat released to the hot reservoir is $Q_1 = Q_2 + W$.
Substituting $Q_2 = \frac{W}{4}$, we get $Q_1 = \frac{W}{4} + W = \frac{5W}{4}$.
Given $Q_1 = 250 \, J$, we have $250 = \frac{5W}{4}$.
Solving for $W$, $W = 250 \times \frac{4}{5} = 200 \, J$.

(C) The efficiency of a Carnot engine is given by $\eta = 1/5$.
When this engine is used as a refrigerator,its coefficient of performance $\beta$ is related to the efficiency $\eta$ by the formula: $\beta = \frac{1 - \eta}{\eta}$.
Substituting the value of $\eta$: $\beta = \frac{1 - 1/5}{1/5} = \frac{4/5}{1/5} = 4$.
The coefficient of performance is also defined as the ratio of heat absorbed from the cold reservoir $(Q)$ to the work done on the system $(W)$: $\beta = \frac{Q}{W}$.
Given $W = 50 \ J$,we have $4 = \frac{Q}{50}$.
Therefore,$Q = 4 \times 50 = 200 \ J$.

(B) refrigerator is a device that extracts heat from a low-temperature reservoir (the inside) and transfers it to a high-temperature reservoir (the room) by performing work on the system.
When the door of an operating refrigerator is kept open,the refrigerator continuously extracts heat from the room and releases it back into the room along with the heat generated by the work done by the compressor.
Since the compressor consumes electrical energy and converts it into heat,the total heat rejected into the room is greater than the heat extracted from it.
Therefore,the net effect is an increase in the total thermal energy of the room,causing the room temperature to increase.

(C) For an ideal heat engine,efficiency $\eta$ is given by:
$\eta = \frac{W}{Q_1} = \frac{Q_1 - Q_2}{Q_1} = 1 - \frac{Q_2}{Q_1} = 1 - \frac{T_2}{T_1}$
From this,we get:
$\frac{T_2}{T_1} = 1 - \eta$
$\frac{T_1}{T_2} = \frac{1}{1 - \eta} \quad \dots (i)$
When the heat engine is operated in the backward direction,it acts as a refrigerator. The coefficient of performance $\beta$ is given by:
$\beta = \frac{Q_2}{W} = \frac{Q_2}{Q_1 - Q_2} = \frac{T_2}{T_1 - T_2}$
Dividing the numerator and denominator by $T_2$:
$\beta = \frac{1}{\frac{T_1}{T_2} - 1}$
Substituting the value from equation $(i)$:
$\beta = \frac{1}{\frac{1}{1 - \eta} - 1} = \frac{1}{\frac{1 - (1 - \eta)}{1 - \eta}} = \frac{1 - \eta}{\eta} = \frac{1}{\eta} - 1$

(A) The coefficient of performance $(\beta)$ of a refrigerator is defined as the ratio of the heat extracted from the cold reservoir $(Q)$ to the work done on the system $(W)$:
$\beta = \frac{Q}{W}$
Given:
$\beta = 5$
$Q = 5000 \,J$
We need to find the electrical energy utilised by the motor, which is equal to the work done $(W)$:
$W = \frac{Q}{\beta}$
$W = \frac{5000 \,J}{5} = 1000 \,J$
Since $1000 \,J = 1 \,kJ$, the electrical energy utilised is $1 \,kJ$.

(B) The coefficient of performance $(COP)$ of a refrigerator is given by $\beta = \frac{T_2}{T_1 - T_2}$.
Here,$T_2 = 0^{\circ} C = 273 \ K$ and $T_1 = 27.3^{\circ} C = 273 + 27.3 = 300.3 \ K \approx 300 \ K$.
Thus,$\beta = \frac{273}{300 - 273} = \frac{273}{27} \approx 10.11$.
Using the relation $\beta = \frac{Q_2}{W}$,where $Q_2$ is the heat extracted to freeze $1 \ g$ of water.
$Q_2 = m \times L_{\text{ice}} = 1 \ g \times 80 \ cal/g = 80 \ cal$.
Converting $Q_2$ to Joules: $Q_2 = 80 \times 4.2 \ J = 336 \ J$.
Therefore,the work done $W = \frac{Q_2}{\beta} = \frac{336}{10.11} \approx 33.23 \ J$.
Given the options,the closest value is $33.6 \ J$ (calculated using $\beta = 10$).
So,$W = 336 / 10 = 33.6 \ J$.

(D) When the door of a refrigerator is opened,the refrigerator extracts heat from the inside and releases it into the room. Additionally,the work done by the compressor to run the refrigerator is also converted into heat and released into the room. Therefore,the net effect is an increase in the room's temperature rather than cooling it. Thus,Assertion $(A)$ is false.
The Reason $(R)$ states that heat flows from a higher temperature body to a lower temperature body,which is a fundamental principle of thermodynamics (Second Law of Thermodynamics). Thus,Reason $(R)$ is true.

(B) The heat extracted from the water to freeze it is $Q_2 = m L = 1 \ g \times 80 \ cal/g = 80 \ cal$.
Converting this to Joules: $Q_2 = 80 \times 4.2 \ J = 336 \ J$.
The coefficient of performance $(K)$ of a refrigerator is given by $K = \frac{Q_2}{W} = \frac{T_2}{T_1 - T_2}$.
Here,$T_2 = 0^{\circ} C = 273 \ K$ and $T_1 = 27.3^{\circ} C = 273 + 27.3 = 300.3 \ K$.
Substituting the values: $\frac{336}{W} = \frac{273}{300.3 - 273} = \frac{273}{27.3} = 10$.
Therefore,$W = \frac{336}{10} = 33.6 \ J$.

(A) The efficiency of a reversible heat engine is given by $\eta = 1 - \frac{T_2}{T_1} = 0.5$.
From this,we find $\frac{T_2}{T_1} = 1 - 0.5 = 0.5$.
The coefficient of performance $(COP)$ of a refrigerator working between the same temperatures is given by $\beta = \frac{T_2}{T_1 - T_2}$.
Dividing the numerator and denominator by $T_1$,we get $\beta = \frac{T_2/T_1}{1 - T_2/T_1}$.
Substituting the value $\frac{T_2}{T_1} = 0.5$,we get $\beta = \frac{0.5}{1 - 0.5} = \frac{0.5}{0.5} = 1$.

(B) The total heat to be removed $(Q)$ consists of two parts: cooling the water from $30^{\circ}C$ to $0^{\circ}C$ and freezing the water at $0^{\circ}C$ into ice.
$1$. Heat to cool water: $Q_1 = m \cdot c \cdot \Delta T = 15 \ kg \times 4200 \ J/(kg \cdot K) \times 30 \ K = 1,890,000 \ J$.
$2$. Heat to freeze water: $Q_2 = m \cdot L_f = 15 \ kg \times 3.36 \times 10^5 \ J/kg = 5,040,000 \ J$.
Total heat $Q = Q_1 + Q_2 = 1,890,000 + 5,040,000 = 6,930,000 \ J$.
Time $t = 1 \ hour = 3600 \ s$.
Power $P = Q / t = 6,930,000 / 3600 = 1925 \ W$.