The specific heat capacity of a metal at low temperature \[(T)\] is given as \[{C_p}(KJ{K^{ - 1}}) = 32{\left( {\dfrac{T}{{400}}} \right)^2}\]. A \[100gm\] vessel of this metals to be cooled from \[{20^o}K\] to \[{4^o}K\] by a special refrigerator operating at room temperature\[({27^o}C)\]. The amount of work required to cool the vessel is:
A) Equal to \[0.002KJ\]
B) Greater than \[0.148KJ\]
C ) Between \[0.148KJ\] and \[0.028KJ\]
D) Less than \[0.028KJ\]
Answer
298.2k+ views
Hint:This is an isothermal process, so we consider \[U = 0\]. Now we use the first law of thermodynamics to calculate the amount of work done.
We substitute the value of \[Q\] after integrating \[dQ = m{C_P}dt\].
The integration is performed within the intervals over which the cooling takes place.
Complete step-by-step answer:
The given process is an isothermal process, therefore, the temperature remains constant, and hence we consider the room temperature which is given as \[({27^o}C)\].
Since the temperature is constant, we can consider change in internal energy to be zero.
Therefore, \[U = 0\].
Using First law of thermodynamics, as we know:
\[\Delta U = Q - W\]
Therefore, we obtain:
\[Q = W\]
where,
\[Q = \]Heat added
\[W = \]Work done by the system.
We need to find the value of \[W\] in the question.
We know:
\[dQ = m{C_P}dt\]
Where:
\[m\] is the mass of substance
\[{C_p}\] is specific heat capacity at constant pressure
\[dt\] is a change in temperature.
Therefore, we integrate \[dQ\] to get \[Q\] which is equal to \[W\].
\[W = Q = \smallint m{C_P}dt\]
Putting the value of \[{C_P}\] as given in the question, we obtain:
\[W = Q = \smallint m \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT\]
It is given that the body cools from \[{20^o}K\]to \[{4^o}K\], so we integrate within this interval.
Thus:
\[W = Q = \smallint _{20}^4m \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT\]
Mass of the vessel is given as \[100gm\], thus we take \[m = 0.1Kg\]
Thus,
\[W = Q = \smallint _{20}^40.1 \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT\]
Thus after solving the equation, we obtain:
\[W = Q = 0.002KJ\].
Hence, option (A) is correct.
Note:Isothermal process is referred to as the process of thermodynamics which takes place at a constant temperature and thus exchange of energy takes place very slowly. This is the reason as a result of which thermal equilibrium is maintained.
Therefore, there is no heat transferred between the system and the surrounding.
We substitute the value of \[Q\] after integrating \[dQ = m{C_P}dt\].
The integration is performed within the intervals over which the cooling takes place.
Complete step-by-step answer:
The given process is an isothermal process, therefore, the temperature remains constant, and hence we consider the room temperature which is given as \[({27^o}C)\].
Since the temperature is constant, we can consider change in internal energy to be zero.
Therefore, \[U = 0\].
Using First law of thermodynamics, as we know:
\[\Delta U = Q - W\]
Therefore, we obtain:
\[Q = W\]
where,
\[Q = \]Heat added
\[W = \]Work done by the system.
We need to find the value of \[W\] in the question.
We know:
\[dQ = m{C_P}dt\]
Where:
\[m\] is the mass of substance
\[{C_p}\] is specific heat capacity at constant pressure
\[dt\] is a change in temperature.
Therefore, we integrate \[dQ\] to get \[Q\] which is equal to \[W\].
\[W = Q = \smallint m{C_P}dt\]
Putting the value of \[{C_P}\] as given in the question, we obtain:
\[W = Q = \smallint m \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT\]
It is given that the body cools from \[{20^o}K\]to \[{4^o}K\], so we integrate within this interval.
Thus:
\[W = Q = \smallint _{20}^4m \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT\]
Mass of the vessel is given as \[100gm\], thus we take \[m = 0.1Kg\]
Thus,
\[W = Q = \smallint _{20}^40.1 \times 32{\left( {\dfrac{T}{{400}}} \right)^2} \times dT\]
Thus after solving the equation, we obtain:
\[W = Q = 0.002KJ\].
Hence, option (A) is correct.
Note:Isothermal process is referred to as the process of thermodynamics which takes place at a constant temperature and thus exchange of energy takes place very slowly. This is the reason as a result of which thermal equilibrium is maintained.
Therefore, there is no heat transferred between the system and the surrounding.
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