Question

If the temperature scale is changed from $^\circ C$ to $^\circ F$, the numerical value of specific heat will
A. Increase
B. Decrease
C. Remain unchanged
D. Nothing can be said

Answer 1

Hint: As we all know that specific heat is inversely proportional to the change in temperature and that is basically the difference in temperature and it means that how the boiling and freezing points are place on the temperature scale used and in that way the value of specific heat is calculated.

Complete step by step answer:
Let us consider a body of mass m and specific heat ‘c’ and we want to heat from state 1 to state 2 at a temperature of ${T_{c1}}^\circ C$ to ${T_{c2}}^\circ C$ where these temperatures are the temperatures in Celsius scale.
So we can know that the heat supplied to the body can be written as:
$Q = mc\Delta T$ …… (I)
Here, $\Delta T$ is the change in temperature and $Q$ is the heat supplied.
We will now solve equation (I) further and it will become,
\[Q = mc\left( {{T_{c2}} - {T_{c1}}} \right)\]
$ \Rightarrow c = \dfrac{Q}{{m\left( {{T_{c2}} - {T_{c1}}} \right)}}$ ….. (II)
Now we know that the relation between the Celsius scale and Fahrenheit scale is,
$T(^\circ F) = \dfrac{9}{5}T(^\circ C) + 32$ …… (III)
Here, $T(^\circ F)$ is the temperature in the Fahrenheit scale, and $T(^\circ C)$ is the temperature in the Celsius scale.
Let us suppose,
${T_{F1}}$= Temperature of state 1 in the Fahrenheit scale
${T_{F2}}$= Temperature of state 2 in the Fahrenheit scale.
${T_{c1}}$= Temperature of state 1 in Celsius scale.
${T_{c2}}$= Temperature of state 2 in Celsius scale.
Now we can write these all temperatures using equation (III) as,
$ \Rightarrow {T_{F1}} = \dfrac{9}{5}{T_{c1}} + 32$
$ \Rightarrow {T_{F2}} = \dfrac{9}{5}{T_{c2}} + 32$
Now we will write the change in temperature w.r.t to Fahrenheit scale
$\Delta {T_F} = {T_{F2}} - {T_{F1}}$ …… (IV)
Here, $\Delta {T_F}$ is the change in temperature in the Fahrenheit scale.
Now we will substitute the values of ${T_{F1}}$ and ${T_{F2}}$ in equation (IV) and we will get,
$\Delta {T_F} = \left( {\dfrac{9}{5}{T_{c2}} + 32} \right) - \left( {\dfrac{9}{5}{T_{c1}} + 32} \right)$
\[ \Rightarrow \Delta {T_F} = \left( {\dfrac{9}{5}{T_{c2}}} \right) - \left( {\dfrac{9}{5}{T_{c1}}} \right)\]
\[ \Rightarrow \Delta {T_F} = \dfrac{9}{5}\left( {{T_{c2}} - {T_{c1}}} \right)\]
We will now represent the specific heat ${c_F}$ in terms of the Fahrenheit scale as taking reference from equation (II). So, we will get,
\[ \Rightarrow c = \dfrac{Q}{{m\left( {{T_{F2}} - {T_{F1}}} \right)}}\] …… (V)
We will now substitute $\Delta {T_F} = {T_{F2}} - {T_{F1}}$ in equation (V) to simplify the value of ${c_F}$.
$ \Rightarrow {c_F} = \dfrac{Q}{{\dfrac{9}{5}m\left( {{T_{c2}} - {T_{c1}}} \right)}}$ …… (VI)
We will now divide equation (II) by equation (VI) to find the relation between $c$ and ${c_F}$.
\[
   \Rightarrow \dfrac{c}{{{c_F}}} = \dfrac{{\dfrac{Q}{{m\left( {{T_{c2}} - {T_{c1}}} \right)}}}}{{\dfrac{Q}{{\dfrac{9}{5}m\left( {{T_{c2}} - {T_{c1}}} \right)}}}} \\
   \Rightarrow \dfrac{c}{{{c_F}}} = \dfrac{9}{5} \\
  \therefore {c_F} = \dfrac{5}{9}c \\
 \]
We can see that this clearly mentions that ${c_F} < c$. So, we say that If the temperature scale is changed from $^\circ C$ to $^\circ F$, the numerical value of specific heat will decrease.

$\therefore$ The correct option is (B).

Note:
We know that 1 degree Celsius is equal to 33.8 degrees Fahrenheit and the value of specific heat decreases when we change the scale from the Celsius scale to the Fahrenheit scale. Suppose if the Celsius scale is changed to the Rankine scale, then also the specific heat decreases as 1 degree Celsius is equal to 493.47 degrees Rankine.