Answer
Verified
343.8k+ views
Hint: The working substance absorbs some amount of heat from the sink which is at the lower temperature, then W is the work which is done on the working substance and then some amount of heat rejected by the working substance to the surroundings. This is the basic principle of refrigerators.
Complete step by step answer:
Given inside temperature (sink temperature), $T_{2}=275 K$
Outside temperature (source temperature), $T_{1}=300K$
Work done = $1 J$
R is the refrigerator.
Assume: Refrigerator is ideal and reversible. So, we can write:
$\dfrac{ Q_{1}}{ Q_{2}} = \dfrac{ T_{1}}{ T_{2}}$
We have to find the heat lost to the surroundings, $ Q_{1}$.
Coefficient of performance -
$B = \dfrac{ Q_{2}}{W}$
Now, we will write B in terms of the temperatures.
$B = \dfrac{ T_{2}}{ T_{1}- T_{2}}$
$B = \dfrac{ 275}{ 300- 275} =11$
By using, formula of coefficient of performance, we can evaluate: -
$B = \dfrac{ Q_{2}}{W}$
Now, we will put the value of W.
$10 = \dfrac{ Q_{2}}{1}$
$ Q_{2} = 10 j$
By using the value of work, we can calculate $ Q_{1}$.
$W = Q_{1} - Q _{2}$
$Q_{1} = W + Q _{2}$
$Q_{1} = 1 + 10 = 11 J$
So, the heat delivered to the surroundings = $11 J$.
So, the correct answer is “Option b”.
Note: The heat flow follows the first law of thermodynamics. Work has to be done, otherwise heat will not flow from low temperature to high temperature and violates the second law of thermodynamics. In the refrigerator, heat is rejected to surroundings and entropy is increased.
Complete step by step answer:
Given inside temperature (sink temperature), $T_{2}=275 K$
Outside temperature (source temperature), $T_{1}=300K$
Work done = $1 J$
R is the refrigerator.
Assume: Refrigerator is ideal and reversible. So, we can write:
$\dfrac{ Q_{1}}{ Q_{2}} = \dfrac{ T_{1}}{ T_{2}}$
We have to find the heat lost to the surroundings, $ Q_{1}$.
Coefficient of performance -
$B = \dfrac{ Q_{2}}{W}$
Now, we will write B in terms of the temperatures.
$B = \dfrac{ T_{2}}{ T_{1}- T_{2}}$
$B = \dfrac{ 275}{ 300- 275} =11$
By using, formula of coefficient of performance, we can evaluate: -
$B = \dfrac{ Q_{2}}{W}$
Now, we will put the value of W.
$10 = \dfrac{ Q_{2}}{1}$
$ Q_{2} = 10 j$
By using the value of work, we can calculate $ Q_{1}$.
$W = Q_{1} - Q _{2}$
$Q_{1} = W + Q _{2}$
$Q_{1} = 1 + 10 = 11 J$
So, the heat delivered to the surroundings = $11 J$.
So, the correct answer is “Option b”.
Note: The heat flow follows the first law of thermodynamics. Work has to be done, otherwise heat will not flow from low temperature to high temperature and violates the second law of thermodynamics. In the refrigerator, heat is rejected to surroundings and entropy is increased.
Recently Updated Pages
Basicity of sulphurous acid and sulphuric acid are
Assertion The resistivity of a semiconductor increases class 13 physics CBSE
Three beakers labelled as A B and C each containing 25 mL of water were taken A small amount of NaOH anhydrous CuSO4 and NaCl were added to the beakers A B and C respectively It was observed that there was an increase in the temperature of the solutions contained in beakers A and B whereas in case of beaker C the temperature of the solution falls Which one of the following statements isarecorrect i In beakers A and B exothermic process has occurred ii In beakers A and B endothermic process has occurred iii In beaker C exothermic process has occurred iv In beaker C endothermic process has occurred
The branch of science which deals with nature and natural class 10 physics CBSE
What is the stopping potential when the metal with class 12 physics JEE_Main
The momentum of a photon is 2 times 10 16gm cmsec Its class 12 physics JEE_Main