Question

2. Caloriemetry
If temperature scale is charged from ${}^{\circ }Ct{o}^{\circ }F$ the numerical value of specific heat will

• A. increases
• B. decreases
• C. Remain uncharged
• D. Become zero

Answer 1

The correct option is B decreases

$\text{Solution}:$

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.

Solving equation (I) further it will become,

$Q=mc\left(T_{c2}-T_{c1}\right)⟹c=\frac{Q}{m\left(T_{c2}-T_{c1}\right)}$ ….. (II)

Now the relation between the Celsius scale and Fahrenheit scale is,

$T{(}^{\circ }F)=\frac{9}{5}T{(}^{\circ }C)+32$ …… (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,

$⟹T_{F1}=\frac{9}{5}{T}_{c1}+32$

$⟹T_{F2}=\frac{9}{5}{T}_{c2}+32$

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,

$\Delta T_F=\left(\frac{9}{5}{T}_{c2}+32\right)-\left(\frac{9}{5}{T}_{c1}+32\right)$

$⟹\Delta T_F=\left(\frac{9}{5}{T}_{c2}\right)-\left(\frac{9}{5}{T}_{c1}\right)$

$⟹\Delta T_F=\frac{9}{5}\left(T_{c2}-T_{c1}\right)$

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,

$⟹c=\frac{Q}{m\left(T_{F2}-T_{F1}\right)}$ …… (V)

We will now substitute $\Delta T_F=T_{F2}-T_{F1}$ in equation (V) to simplify the value of $c_F$

$⟹c_F=\frac{Q}{\frac{9}{5}m\left(T_{c2}-T_{c1}\right)}$…… (VI)

We will now divide equation (II) by equation (VI) to find the relation between c and $c_F$

$⟹\frac{c}{c_F}=\frac{\frac{Q}{m\left(T_{c2}-T_{c1}\right)}}{\frac{Q}{\frac{9}{5}m\left(T_{c2}-T_{c1}\right)}}$

$⟹\frac{c}{c_F}=\frac{9}{5}$

$\therefore c_F=\frac{5}{9}c$

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.

$\text{Hence B is the correct option}$